158. Very nice and difficult limits.
by Virgil Nicula, Oct 16, 2010, 3:41 PM
PP1. Let 
be a sequence for which 
Prove that ![$ \lim_{n\to\infty}\ \left(\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right)=\frac {el}{2}$](//latex.artofproblemsolving.com/9/2/8/928f8ccd071c6d6b52727782d9b23713c3ccbb12.png)
.
Particular case. For
obtain ![$ \lim_{n\to\infty}\ \left(\frac {(n+1)^2}{\sqrt[n+1]{(n+1)!}}-\frac {n^2}{\sqrt[n]{n!}}\right)=e$](//latex.artofproblemsolving.com/8/2/b/82b158774d426618123c31b29e7c30e67ae86741.png)
(D.M. Batinetu-Giurgiu's sequence).
1. Preliminary.![$\left\{\begin{array}{ccc}\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac {x_{n+1}-x_n}{(n+1)^2-n^2}=\lim_{n\to\infty}\ \frac {x_{n+1}-x_n}n\cdot\frac n{2n+1}=\frac 12\cdot l\ \stackrel{\mathrm{(ST)}}{\implies}\ \lim_{n\to\infty}\ \frac {x_n}{n^2}=\frac l2\ } \\\ \\\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac {x_{n+1}}{(n+1)^2}\cdot\left(\frac {n+1}n\right)^2\cdot\frac {n^2}{x_n}=\frac l2\cdot 1^2\cdot \frac 2l\ \implies\ \lim_{n\to\infty}\ \frac {x_{n+1}}{x_n}=1\ } \\\ \\\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac {x_n}{n^2}\cdot n^2=\frac l2\cdot \infty\ \implies\ \lim_{n\to\infty}\ x_n=\infty\ } \\\ \\\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac n{\sqrt[n]{n!}}=\lim_{n\to\infty}\ \left(1+\frac 1n\right)^n=e\ \ ;\ \ \lim_{x\to 0}\ \frac {\ln (1+x)}x=\lim_{x\to 0}\ \frac {e^x-1}x=1\ }\end{array}\right\|$](//latex.artofproblemsolving.com/9/c/d/9cd6f1264ad45af932b65e9716032c2c53d956bd.png)
.
2. Proof of the proposed problem.
![$\lim_{n\to\infty}\ \left(\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right)=$](//latex.artofproblemsolving.com/e/5/b/e5b828b6a2580b2e4da621b4f1e01c434ab2cf1d.png)
![$\lim_{n\to\infty}\ \frac {n+1}{\sqrt[n+1]{(n+1)!}}\cdot\frac n{\sqrt[n]{n!}}\cdot\frac n{n+1}\cdot\frac 1{n^2}\cdot\left(x_{n+1}\sqrt[n]{n!}-x_n\sqrt[n+1]{(n+1)!}\right)=$](//latex.artofproblemsolving.com/2/0/6/206b463dbde13bfea8ff22c6fb122beff92800f9.png)
![$=e^2\cdot\lim_{n\to\infty}\ \frac {x_n}{n^2}\cdot\frac {\sqrt[n+1]{(n+1)!}}{n+1}\cdot\frac {n+1}n\cdot n\cdot\left(\frac {x_{n+1}\sqrt[n]{n!}}{x_n\sqrt[n+1]{(n+1)!}}-1\right)=$](//latex.artofproblemsolving.com/0/e/1/0e103acca42b4e59c0fa527b043a6ffde36a590d.png)
, where
![$b_n=\frac {x_{n+1}\sqrt[n]{n!}}{x_n\sqrt[n+1]{(n+1)!}}$](//latex.artofproblemsolving.com/3/6/b/36ba21dcc8574cd694119ba95c0ebeaf1afa9d4c.png)
and 
. Therefore, it remains to prove that : 
. Indeed,
and 
. On the other hand we have :
and![$\lim_{n\to\infty}\ \frac {\left[\ln (n+1)!-(n+1)\ln (n+2)\right]-\left[\ln n!-n\ln (n+1)\right]}{(n+2)-(n+1)}=$](//latex.artofproblemsolving.com/b/9/c/b9c1615b51a7d363e0ac186f3892563587fd92cc.png)
.
Now, returning in the chain of equalities
with the relations 
& 
we obtain: 
, as desired .
PP2. Prove that :
, where 
and 
.
Proof. We will use the following known inequalities :
. Thus, from
obtain the inequality :
, which multiplied by 
becomes : ![$\boxed{\ \frac {\left(\ln n-1\right)\cdot\ln n}{H_n^2}\ <\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}\ <\ \frac {\left[n\cdot\left(\ln n-1\right)+\ln n\right]\cdot \ln n}{n\cdot H_n^2}\ }\ ,\ \forall\ n\ge 7$](//latex.artofproblemsolving.com/2/9/9/299f8cee8082032a705496b8abbd391bb676c937.png)
.
On the other hand, from the inequalities we can weaken the last chain of inequalities and finally obtain that :
![$\boxed{\ \frac {\left(\ln n-1\right)\cdot \ln n }{\left[1+\ln (n+1)\right]^2}\ <\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}\ <\ \frac {n\cdot\left(\ln n-1\right)+\ln n}{n\cdot \ln n}\ }\ \ ,\ \forall\ n\ge 7\ \ (\ast)$](//latex.artofproblemsolving.com/7/2/7/727ce7fe3ed85323d6e97f1e4cb4de319f5c0b59.png)
.
Since![$\left\{\begin{array}{lllll}
\lim\limits_{n\to\infty}\ \frac {\left(\ln n-1\right)\cdot \ln n }{\left[1+\ln (n+1)\right]^2}=\lim\limits_{n\to\infty}\ \frac {1-\frac {\ln n}n}{\left[\frac {1+\ln (n+1)}{\ln n}\right]^2}=1 \\\\\
\lim\limits_{n\to\infty}\ \frac {n\cdot\left(\ln n-1\right)+\ln n}{n\cdot \ln n}=\lim\limits_{n\to\infty}\ \left[1-\frac 1{\ln n}+\frac 1n\right]=1\end{array}\right\|$](//latex.artofproblemsolving.com/5/a/3/5a30d9ffa75e598b30001c2f48e6a7f4af5db585.png)
.
Therefore,
, so we are done.
PP3. Let
and 
for 
. Prove that 
, 
and find 
.
Proof. It is obvious that the sequence
is decreasing . By induction one can easily prove that : 
.
Thus,![$x_n\to l\in [0,1]$](//latex.artofproblemsolving.com/c/1/4/c140a600e40a0d6cedded0d9cdd47f8f7b806d75.png)
and by passing to limit in the recurrence relation we get : 
i.e. 
.
On the other hand,
.
Accordingly, by the Stolz-Cesaro theorem we obtain that :
. Hence, 
. But : ![$\lim_{n\to\infty}\ \frac {\left[\frac 1{x_{n+1}}-(n+1)\right]-\left[\frac 1{x_n}-n\right]}{\ln (n+1)-\ln n}=\lim_{n\to\infty}\ \frac {\frac 1{x_{n+1}}-\frac 1{x_n}-1}{\ln\left(1+\frac 1n\right)}$](//latex.artofproblemsolving.com/5/1/5/5150099f01043316e3abc9500badb507586debd8.png)
![$=\lim_{n\to\infty}\ n\cdot\left[\frac 1{x_n\cdot (1-x_n)}-\frac 1{x_n}-1\right]=$](//latex.artofproblemsolving.com/6/c/d/6cdd064c45a79f90e0348e4ed53b232367d46f92.png)
. Therefore,
using again the Stolz-Cesaro theorem we obtain the required limit :
.
PP4. Prove that![$\lim_{n\to\infty}\ \left[\frac {\left(1+\frac 1n\right)^n}{e}\right]^n=\frac {1}{\sqrt e}$](//latex.artofproblemsolving.com/0/4/b/04b6a8d525ac52448febf3fb6b4b5a950e5947e9.png)
.
Proof.![$l\equiv \lim_{n\to\infty}\ \left[\frac {\left(1+\frac 1n\right)^n}{e}\right]^n=e^L$](//latex.artofproblemsolving.com/a/4/1/a4176daeab00b95c1c9cbe5f484340f89c06818e.png)
, where ![$L=\frac 1e\cdot \lim_{n\to\infty}\ n\left[\left(1+\frac 1n\right)^n-e\right]\ \stackrel{(*)}{=}\ \frac 1e\cdot\left(-\frac e2\right)=-\frac 12$](//latex.artofproblemsolving.com/1/7/e/17e28a39a99d51f8fa7f743a719728d0f5d20837.png)
. In conclusion, 
.
=====================================================================================================================
![$\blacktriangleright\ (*)\ \lim_{n\to\infty}\ n\cdot\left[\left(1+\frac 1n\right)^n-\text{e}\right]=$](//latex.artofproblemsolving.com/a/7/7/a779b24fd207dce454f8c379fc3d9d89e46ba570.png)
![$\lim_{n\to\infty}\ n\text{e}\cdot\left[\frac 1{\text{e}}\left(1+\frac 1n\right)^n-1\right]=$](//latex.artofproblemsolving.com/9/3/a/93a2b796b8b378e64a713974537d859ada876325.png)
![$\text{e}\cdot\lim_{n\to\infty}\ n\cdot\left[\text{e}^{n\ln\left(1+\frac 1n\right)-1}-1\right]=$](//latex.artofproblemsolving.com/5/2/3/523b68a4120f509c795a9531f14af311f97baabb.png)
![$\text{e}\cdot\lim_{n\to\infty}\ n\cdot\left[n\ln\left(1+\frac 1n\right)-1\right]=$](//latex.artofproblemsolving.com/8/1/f/81f53ee73d46fa50bd7bd5c45c5351ce2f5cd212.png)
![$-\text{e}\cdot\lim_{n\to\infty}\ \left[n-n^2\ln\left(1+\frac 1n\right)\right]=$](//latex.artofproblemsolving.com/a/e/a/aeaf7394f7caeb8171b0b2ac2e5c0e11625fe168.png)
, where 
and 
.
Proof.![$\lim_{x\searrow 0}\left[\frac {(1+x)^{\frac 1x}}{e}\right]^{\frac 1x}=e^l$](//latex.artofproblemsolving.com/5/c/2/5c261d00e36d293b1a07d406d3c8f3ccca9d65df.png)
, where ![$l=\lim_{x\searrow 0}\frac 1x\cdot \left[\frac {(1+x)^{\frac 1x}}{e}-1\right]=$](//latex.artofproblemsolving.com/6/9/a/69a0122806edfc89bc1c5f7fbc1fb6f313f6cdb6.png)
![$\boxed{\ \lim_{x\searrow 0}\left[\frac {(1+x)^{\frac 1x}}{e}\right]^{\frac 1x}=\frac {1}{\sqrt e}\ }$](//latex.artofproblemsolving.com/e/9/8/e98a82a6f9bb7021a83bbd4274497d0a3e27629e.png)
.
Remark. I"used the remarkable limits
and 
.
PP5. Prove that
, where 
.
Proof. Prove easily that
, where 
and 
. Using the well-known limits 
(Euler's number) and
obtaim that 
.
Remark. Ascertain analogously that
.
PP6. Let
and 
, for any 
. Prove that 
.
Proof. The recurrence relation can be written as
and 
. Observe that 
, so 
.
But
. From Stolz-Cesaro theorem obtain that : 
.
PP7.
![${\color{white}{.}}\implies\lim_{n\to\infty}\, \left(\sqrt[n+1]{\frac {x_{n+1}}{y_{n+1}}}-\sqrt[n]{\frac {x_n}{y_n}}\ \right)=\frac a{b\cdot \text{e}}$](//latex.artofproblemsolving.com/0/9/8/098820d616b6d6d937025fc6d0620071d272a738.png)
Remark. In particular, for
and 
we obtain Traian Lalescu's sequence i.e. ![$\lim_{n\to\infty}\, \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\frac 1{\text{e}}$](//latex.artofproblemsolving.com/2/3/3/233f1a07338a0b165c74280c32cb813a8df4e177.png)
Proof. Let us denote :![$a_n=\sqrt[n]{\frac {x_n}{y_n}}$](//latex.artofproblemsolving.com/6/a/a/6aa5b61299c7836dad2ffeba7c01d65b8576d53b.png)
. One can easily show, by Cauchy-d'Alembert criterion that : 
. Moreover, we can obtain that :
![$\left\{\ \begin{array}{ccc}
\lim\limits_{n\to\infty}\, \frac {a_{n+1}}{a_n}=\lim\limits_{n\to\infty}\, \left[\frac {a_{n+1}}{n+1}\cdot\frac n{a_n}\cdot\frac {n+1}n\right]=\frac a{b\text{e}}\cdot\frac {b\text{e}}a\cdot 1=1 \\ \\
\lim\limits_{n\to\infty}\, \left(\frac {a_{n+1}}{a_n}\right)^n=\lim\limits_{n\to\infty}\, \left(\frac {x_{n+1}}{n^2x_n}\cdot\frac {ny_n}{y_{n+1}}\cdot n\sqrt[n]{\frac {y_n}{x_n}}\right)^{\frac n{n+1}}=\frac ab\cdot\frac {b\text{e}}a=\text{e}\ \end{array}\right\|$](//latex.artofproblemsolving.com/c/3/d/c3d0161b9a4128fdb4840f990d1252185668a22f.png)
. On the other hand, note that : ![$ \lim_{n\to\infty}\, \left(\sqrt[n+1]{\frac {x_{n+1}}{y_{n+1}}}-\sqrt[n]{\frac {x_n}{y_n}}\ \right)=$](//latex.artofproblemsolving.com/8/e/3/8e32be963a26b86315bf25a4ed9a652232cbc4bc.png)
![$\lim_{n\to\infty}\, \left(a_{n+1}-a_n\right)=\lim_{n\to\infty}\, \left[a_n\cdot\left(\frac {a_{n+1}}{a_n}-1\right)\right]=$](//latex.artofproblemsolving.com/c/c/4/cc4bf7d706a0f42f16f412338f09c5a227dec943.png)
![$\lim_{n\to\infty}\ \left[\frac {a_n}n\cdot\frac {\text{e}^{\ln\left(\frac {a_{n+1}}{a_n}\right)}-1}{\ln\left(\frac {a_{n+1}}{a_n}\right)}\cdot\ln\left(\frac {a_{n+1}}{a_n}\right)^n\right]=\frac a{b\cdot\text{e}}\cdot 1\cdot\ln\text{e}=\frac a{b\cdot\text{e}}$](//latex.artofproblemsolving.com/b/d/b/bdbaac9a094819832defe12e9bd66ff2843c1bf1.png)
, as required.
PP8. Prove that
and 
, where 
.
Proof. One can easily prove by induction that the sequence
has positive terms. On the other hand, by Bernoulli's inequality one gets: 
- note that 
- so the given sequence it is decreasing, thus convergent and let 
be its limit. By passing to limit in the recurrent
relation we obtain:
and differentiating with respect to 
both sides of the latter equality we now have: 
whence the desired
limit :
. For the second part of our problem let's note that the required limit, say 
, can be written as : 
so this is an indetermination
case
- because is monotonously decreasing and its limit is equal to 
then 
is monotonously increasing to 
. Now we need to find the
limit
and in case this is finite then, by Stolz-Cesaro theorem, we will conclude that 
. Observe that : 
![$\lim_{n\to\infty}\, \frac {x_n\left[\left(1+\frac {x_n}a\right)^a-1\right]}{x_n-\left(1+\frac {x_n}a\right)^a+1}=$](//latex.artofproblemsolving.com/d/d/e/dde0d4ea9703a5043ab058e6184725c597d19365.png)
![$\lim_{n\to\infty}\left[ \frac {\left( 1+\frac {x_n}a\right )^{a-1}}{\frac {x_n}a\cdot\frac {x_n^2}a\cdot\frac 1{1+x_n}-\left(1+\frac {x_n}a\right)^a} \right]=$](//latex.artofproblemsolving.com/9/d/e/9def6b1eedfb72fb1b5ba40fccd03efccf7f1834.png)
. But
so
and accordingly 
.
PP9. Is well-known that
, where 
. Prove that 
and 
.
Proof.
, where 
. 
, where 
.
PP10. Prove that
, where 
for 
.
Proof.
,
where
and 
. In conclusion, 
and 
.
PP11. Prove that![$\lim_{n\to\infty}\sqrt [n]{\frac {C_{3n}^n}{C_{2n}^n}}=$](//latex.artofproblemsolving.com/e/8/2/e82e41a123fd955a880f6fb1e38f08bca9dd0b9f.png)
. In general, ![$\{p,q\}\subset\mathbb N\ ,\ \ p>q\ge 2\ \implies\ \lim_{n\to\infty}\ \sqrt [n]{\frac {C_{pn}^n}{C_{qn}^n}}=\frac {p^p(q-1)^{q-1}}{q^q(p-1)^{p-1}}$](//latex.artofproblemsolving.com/6/8/2/682fa9d40c1cb1b3cfa98c39ac35759dd9163050.png)
.
Proof.![$a_n=\sqrt [n]{x_n}$](//latex.artofproblemsolving.com/d/b/b/dbb2c2b54872b62a5d0f7a99e4f2361f4518f30a.png)
, 
. Since 
, from an well-known property obtain that and ![$\boxed{a_n=\sqrt [n]{x_n}\rightarrow \frac {27}{16}}$](//latex.artofproblemsolving.com/6/d/8/6d804daba89e491919b538c8bac8921d22063fac.png)
.
PP12. Find![$\lim_{n\to\infty} n^{b-a}\cdot\sqrt[n]{\frac{(an)!}{(bn)!}}$](//latex.artofproblemsolving.com/c/8/6/c86a833c5211318cd03746962e95463076b56249.png)
, where 
.
Proof. If![$a_n=\sqrt[n]{b_n}$](//latex.artofproblemsolving.com/8/a/7/8a7c7e1a52ba6edbce28b8432a61c94fdf475065.png)
, where 
, then 
and
. In conclusion, using the well-known property ![$\frac {b_{n+1}}{b_n}\rightarrow l\implies \sqrt [n]{b_n}\rightarrow l$](//latex.artofproblemsolving.com/c/7/c/c7c1eaa4265d234bfad345eb41eeb102e856a915.png)
, obtain that 
.
PP13. Prove that![$\boxed{\, L=\lim_{x\to 0}\, \frac {\left[\arcsin\left(\frac {\sin x}x\right)-\arcsin\left(1\right)\right]^2}{\arctan\left(\frac {\tan x}x\right)-\arctan\left(1\right)}=2\, }$](//latex.artofproblemsolving.com/9/2/a/92aaaa954ee9b1119c9ecbfad4d22a5d4c619ad0.png)
.
Proof.
as well as: 
. Thus,
![$L=\lim_{x\to 0}\, \left[\left(\frac {-\arcsin\left(\sqrt {1-\left(\frac {\sin x}x\right)^2}\right)}{\sqrt{1-\left(\frac {\sin x}x\right)^2}}\right)^2\cdot\frac {\frac {\tan x-x}{\tan x+x}}{\arctan\left(\frac {\tan x-x}{\tan x+x}\right)}\cdot\frac {1-\left(\frac {\sin x}x\right)^2}{\frac {\tan x-x}{\tan x+x}}\right]$](//latex.artofproblemsolving.com/7/f/a/7fa4e4d82887f5cc3090fe8e5314857329779044.png)
whence, by taking into account that
, our limit reduces to computing 
.
On the other hand, one can easily prove that
. Now it is easy to see that the desired limit can be conveniently written as
![$L=\lim_{x\to 0}\, \left[\frac {x-\sin x}{x^3}\cdot\frac {x^3}{\tan x-x}\cdot\frac {x+\sin x}x\cdot\frac {x+\tan x}x\right]$](//latex.artofproblemsolving.com/1/f/2/1f2e638dacd2a3278140c3ca43e37b00d40ff149.png)
which finally equals 
because
. Here is another related limit, which is equally interesting ![$\boxed{\, L_1=\lim_{x\to 0}\, \frac {\left[\arcsin\left(\frac {\tan x}x\right)-\arcsin\left(1\right)\right]^2}{\arctan\left(\frac {\sin x}x\right)-\arctan\left(1\right)}=8\, }$](//latex.artofproblemsolving.com/b/8/d/b8d6c6f7ec573a34fead37c6dc212931cd332c8b.png)
.
PP14. Calculate the limit![$\lim_{x\to\infty}x\left[\left(\frac {x}{x+1}\right)^x-\frac 1e\right]$](//latex.artofproblemsolving.com/7/6/1/761373ecaf4d7c5ec1af26cde3ec66143b875a6a.png)
.
Proof.![$\lim_{x\to\infty}x\left[\left(\frac {x}{x+1}\right)^x-\frac 1e\right]=$](//latex.artofproblemsolving.com/7/4/8/74885a576e3a02fa8525554bf013568b7546fc0a.png)
![$\frac 1e\cdot\lim_{x\to\infty}x\left[e\left(\frac {x}{x+1}\right)^x-1\right]\stackrel{(*)}{=}$](//latex.artofproblemsolving.com/b/d/9/bd9c2621430cb3fde67518efea9360cfcbbe9b58.png)
![$\frac 1e\cdot\lim_{x\to\infty}x\ln\left[e\left( \frac {x}{x+1}\right)^x\right]=$](//latex.artofproblemsolving.com/5/d/f/5dfa2250a9f5be78a3a00918f1c7e894662cee57.png)
![$\frac 1e\cdot\lim_{x\to\infty}x^2\left[\frac 1x+\ln x- \ln (x+1)\right]=$](//latex.artofproblemsolving.com/5/1/2/512211e532a0d8f5e15c57dd5cbc957ef65ce955.png)
.
Rematk. In upper equality
I used the remarkable limits 
and 
.
PP15. For
find ![$\lim_{x\to \infty}\left[\left(x-x^p\right)^p-x^p\right]$](//latex.artofproblemsolving.com/4/e/3/4e32df89287cf9ef6d84776f07e7f9b841132f6e.png)
.
Proof.![$\lim_{x\to\infty}\left[\left(x-x^p\right)^p-x^p\right]=$](//latex.artofproblemsolving.com/3/2/6/326a66d70ad1572ea30e96ce2c3be5b4ac860894.png)
![$\lim_{x\to\infty}x^p\left[\left(1-x^{p-1}\right)^p-1\right]\stackrel{(*)}{=}$](//latex.artofproblemsolving.com/0/4/a/04a8dc0584e2a052fa2b3707f52f984e5a19c17f.png)
.
Remark. In the equality I used the remarkable limit 
.
PP16. Ascertain the limit![$l=\lim_{n\to\infty}\left[\prod_{k=1}^n\left(1+\frac {2k-1}{2n}\right)\right]^{\frac {1}{2n}}$](//latex.artofproblemsolving.com/8/1/d/81dd8359f6e7373f537b11fcca410e74e91cc1c9.png)
.
Proof.
![$\ln 2-\left|\left[x-\ln (x+1)\right]\right|_0^1=$](//latex.artofproblemsolving.com/d/6/7/d67946b140175de16827dbed6adbeb4e7cddbcff.png)
.
PP17 (without involving l'Hospital rules) Ascertain![$L=\lim_{x\to\infty} x\cdot \left(\sqrt[p]{x^p+ax^{p-1}+bx^{p-2}+\ \ldots}+\sqrt[p]{x^p-ax^{p-1}+cx^{p-2}+\ \ldots}-2x\right)$](//latex.artofproblemsolving.com/a/9/0/a908509bd9d667f2cdc6bbb09a3a97963795f087.png)
.
Proof. Denote![$\left\{\begin{array}{c}
f(x)=\sqrt[p]{x^p+ax^{p-1}+bx^{p-2}+\ \ldots}\\\\
g(x)=\sqrt[p]{x^p-ax^{p-1}+cx^{p-2}+\ \ldots}\end{array}\right\|$](//latex.artofproblemsolving.com/a/1/1/a11062e9f6dedd5627beaa289a42abaf5fc987e8.png)
. Prove easlily that ![$\lim_{x\to\infty}\ \left[f(x)-x\right]=\frac ap$](//latex.artofproblemsolving.com/7/a/6/7a6a7110e9709b58186979a4dc5016ab13d294ff.png)
and ![$\lim_{x\to\infty}\ \left[g(x)-x\right]=-\frac ap$](//latex.artofproblemsolving.com/6/a/c/6ac7a358fd1d9a5ecbf78951e6ab336f115b0810.png)
. Hence we are in the exception case 
. Using the substitution 
obtain that ![$L=\lim_{y\to 0}\frac {1}{y^2}\cdot \left(\sqrt[p]{1+ay+by^2+\ \ldots}+\sqrt[p]{1-ay+cy^2+\ \ldots}-2\right)$](//latex.artofproblemsolving.com/f/e/e/fee7907bdb591f018f8576c347a6d837d8622453.png)
. Denote ![$\left\{\begin{array}{c}
u(y)=\sqrt [p]{1+ay+by^2+\ \ldots}\\\\
v(y)=\sqrt [p]{1-ay+cy^2+\ \ldots}\end{array}\right\|$](//latex.artofproblemsolving.com/1/3/c/13caaab4be1b2431ea5cccd4acba7fcc81e45989.png)
. Observe that 
and 
. Since ![$u(y)+v(y)-2=u(y)v(y)-1-[u(y)-1][v(y)-1]$](//latex.artofproblemsolving.com/7/d/f/7df196b3a725de4ed97c12010f0515f35700349f.png)
obtain that ![$L=\lim_{y\to 0}\ \left[\frac {u(y)v(y)-1}{y^2}-\frac {u(y)-1}{y}\cdot\frac {v(y)-1}{y}\right]=\frac 1p\cdot \left(b+c-a^2\right)+\frac {a^2}{p^2}$](//latex.artofproblemsolving.com/9/b/d/9bd4be9cbc68e124cc1e01ef715f8ce505f7c43f.png)
. In conlusion, 
.
Remark. In the particular case obtain that![$\lim_{x\to\infty}\ x\cdot\left(2x-\sqrt [3]{x^3+x^2+1}-\sqrt [3]{x^3-x^2+1}\right)=\frac 29$](//latex.artofproblemsolving.com/2/d/a/2daf8d4f5897e0adceb4f622e868f859a73e7a84.png)
.
PP18 Prove that![$\lim_{n\to\infty}\ \frac {n\left(\sqrt[n]n-1\right)}{\ln n}=$](//latex.artofproblemsolving.com/4/1/b/41bee6d7db7d19924d3aeb9d9d99c4f6f4a99380.png)
![$1\ ,\ \lim_{n\to\infty}\ n\Big(\sqrt[n]{n}-1 \Big)-\ln{n}$](//latex.artofproblemsolving.com/5/0/a/50a63c0d3e0a7ba95b37658a319b427f12b1d694.png)
![$=0\ ,\ \boxed{\lim_{n\to\infty}\ \frac {n}{\ln^2n}\cdot \Big[n\Big(\sqrt[n]{n}-1 \Big)-\ln{n}\Big]=\frac 12}$](//latex.artofproblemsolving.com/d/5/1/d5153161abd8a05418b385b7dfbe3d4725418ba2.png)
.
Proof. Since
, for 
obtain that 
![$\lim_{n\to\infty}\ \frac {n^2}{\ln^2n}\cdot\left(\sqrt[n]n-1-\frac {\ln n}{n}\right)=\frac 12\implies $](//latex.artofproblemsolving.com/f/e/c/fec43978e79db7cb300ed26ad0502a1d421b17c2.png)
![$\lim_{n\to\infty}\ \frac {n}{\ln^2n}\cdot \Big[n\Big(\sqrt[n]{n}-1 \Big)-\ln{n}\Big]=\frac 12$](//latex.artofproblemsolving.com/5/3/1/531be8aa94cee66a7e2bce1f2b8cc265205f0568.png)
.
PP19. Let
be a sequence defined by 
and 
for any . Prove that 
and find 
.
Proof. From
we have 
. Squaring gives 
or 
with 
and 
. For (a) it suffices to show that 
for all
. This follows immediately from 
and 
. For (b) it suffices to show that 
. Because 
and from (a) , we have
. It is well-known that the harmonic series 
is less than
for some constant 
(Approximate the series by 
to see this). As a result, 
is bounded between 
where 
. Now since
, it is clear that 
which implies the desired result.
PP20. Prove that
, where 
.
Proof. Using the inequality
obtain that 
. Therefore,
.
See here
be a sequence for which 
Prove that ![$ \lim_{n\to\infty}\ \left(\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right)=\frac {el}{2}$](http://latex.artofproblemsolving.com/9/2/8/928f8ccd071c6d6b52727782d9b23713c3ccbb12.png)
.Particular case. For
obtain ![$ \lim_{n\to\infty}\ \left(\frac {(n+1)^2}{\sqrt[n+1]{(n+1)!}}-\frac {n^2}{\sqrt[n]{n!}}\right)=e$](http://latex.artofproblemsolving.com/8/2/b/82b158774d426618123c31b29e7c30e67ae86741.png)
(D.M. Batinetu-Giurgiu's sequence).1. Preliminary.
![$\left\{\begin{array}{ccc}\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac {x_{n+1}-x_n}{(n+1)^2-n^2}=\lim_{n\to\infty}\ \frac {x_{n+1}-x_n}n\cdot\frac n{2n+1}=\frac 12\cdot l\ \stackrel{\mathrm{(ST)}}{\implies}\ \lim_{n\to\infty}\ \frac {x_n}{n^2}=\frac l2\ } \\\ \\\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac {x_{n+1}}{(n+1)^2}\cdot\left(\frac {n+1}n\right)^2\cdot\frac {n^2}{x_n}=\frac l2\cdot 1^2\cdot \frac 2l\ \implies\ \lim_{n\to\infty}\ \frac {x_{n+1}}{x_n}=1\ } \\\ \\\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac {x_n}{n^2}\cdot n^2=\frac l2\cdot \infty\ \implies\ \lim_{n\to\infty}\ x_n=\infty\ } \\\ \\\
{\color[rgb]{0.6,0,0}\blacktriangleright} & \boxed{\ \lim_{n\to\infty}\ \frac n{\sqrt[n]{n!}}=\lim_{n\to\infty}\ \left(1+\frac 1n\right)^n=e\ \ ;\ \ \lim_{x\to 0}\ \frac {\ln (1+x)}x=\lim_{x\to 0}\ \frac {e^x-1}x=1\ }\end{array}\right\|$](http://latex.artofproblemsolving.com/9/c/d/9cd6f1264ad45af932b65e9716032c2c53d956bd.png)
.2. Proof of the proposed problem.
![$\lim_{n\to\infty}\ \left(\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right)=$](http://latex.artofproblemsolving.com/e/5/b/e5b828b6a2580b2e4da621b4f1e01c434ab2cf1d.png)
![$\lim_{n\to\infty}\ \frac {n+1}{\sqrt[n+1]{(n+1)!}}\cdot\frac n{\sqrt[n]{n!}}\cdot\frac n{n+1}\cdot\frac 1{n^2}\cdot\left(x_{n+1}\sqrt[n]{n!}-x_n\sqrt[n+1]{(n+1)!}\right)=$](http://latex.artofproblemsolving.com/2/0/6/206b463dbde13bfea8ff22c6fb122beff92800f9.png)
![$=e^2\cdot\lim_{n\to\infty}\ \frac {x_n}{n^2}\cdot\frac {\sqrt[n+1]{(n+1)!}}{n+1}\cdot\frac {n+1}n\cdot n\cdot\left(\frac {x_{n+1}\sqrt[n]{n!}}{x_n\sqrt[n+1]{(n+1)!}}-1\right)=$](http://latex.artofproblemsolving.com/0/e/1/0e103acca42b4e59c0fa527b043a6ffde36a590d.png)
, where![$b_n=\frac {x_{n+1}\sqrt[n]{n!}}{x_n\sqrt[n+1]{(n+1)!}}$](http://latex.artofproblemsolving.com/3/6/b/36ba21dcc8574cd694119ba95c0ebeaf1afa9d4c.png)
and 
. Therefore, it remains to prove that : 
. Indeed,
and 
. On the other hand we have :
and
![$\lim_{n\to\infty}\ \frac {\left[\ln (n+1)!-(n+1)\ln (n+2)\right]-\left[\ln n!-n\ln (n+1)\right]}{(n+2)-(n+1)}=$](http://latex.artofproblemsolving.com/b/9/c/b9c1615b51a7d363e0ac186f3892563587fd92cc.png)
.Now, returning in the chain of equalities
with the relations 
& 
we obtain: 
, as desired .PP2. Prove that :
, where 
and 
. Proof. We will use the following known inequalities :
. Thus, from obtain the inequality :
, which multiplied by 
becomes : ![$\boxed{\ \frac {\left(\ln n-1\right)\cdot\ln n}{H_n^2}\ <\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}\ <\ \frac {\left[n\cdot\left(\ln n-1\right)+\ln n\right]\cdot \ln n}{n\cdot H_n^2}\ }\ ,\ \forall\ n\ge 7$](http://latex.artofproblemsolving.com/2/9/9/299f8cee8082032a705496b8abbd391bb676c937.png)
.On the other hand, from the inequalities
we can weaken the last chain of inequalities and finally obtain that :![$\boxed{\ \frac {\left(\ln n-1\right)\cdot \ln n }{\left[1+\ln (n+1)\right]^2}\ <\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}\ <\ \frac {n\cdot\left(\ln n-1\right)+\ln n}{n\cdot \ln n}\ }\ \ ,\ \forall\ n\ge 7\ \ (\ast)$](http://latex.artofproblemsolving.com/7/2/7/727ce7fe3ed85323d6e97f1e4cb4de319f5c0b59.png)
.Since
![$\left\{\begin{array}{lllll}
\lim\limits_{n\to\infty}\ \frac {\left(\ln n-1\right)\cdot \ln n }{\left[1+\ln (n+1)\right]^2}=\lim\limits_{n\to\infty}\ \frac {1-\frac {\ln n}n}{\left[\frac {1+\ln (n+1)}{\ln n}\right]^2}=1 \\\\\
\lim\limits_{n\to\infty}\ \frac {n\cdot\left(\ln n-1\right)+\ln n}{n\cdot \ln n}=\lim\limits_{n\to\infty}\ \left[1-\frac 1{\ln n}+\frac 1n\right]=1\end{array}\right\|$](http://latex.artofproblemsolving.com/5/a/3/5a30d9ffa75e598b30001c2f48e6a7f4af5db585.png)
.Therefore,
, so we are done.PP3. Let
and 
for 
. Prove that 
, 
and find 
. Proof. It is obvious that the sequence
is decreasing . By induction one can easily prove that : 
.Thus,
![$x_n\to l\in [0,1]$](http://latex.artofproblemsolving.com/c/1/4/c140a600e40a0d6cedded0d9cdd47f8f7b806d75.png)
and by passing to limit in the recurrence relation we get : 
i.e. 
.On the other hand,
.Accordingly, by the Stolz-Cesaro theorem we obtain that :
. Hence, 
. But : ![$\lim_{n\to\infty}\ \frac {\left[\frac 1{x_{n+1}}-(n+1)\right]-\left[\frac 1{x_n}-n\right]}{\ln (n+1)-\ln n}=\lim_{n\to\infty}\ \frac {\frac 1{x_{n+1}}-\frac 1{x_n}-1}{\ln\left(1+\frac 1n\right)}$](http://latex.artofproblemsolving.com/5/1/5/5150099f01043316e3abc9500badb507586debd8.png)
![$=\lim_{n\to\infty}\ n\cdot\left[\frac 1{x_n\cdot (1-x_n)}-\frac 1{x_n}-1\right]=$](http://latex.artofproblemsolving.com/6/c/d/6cdd064c45a79f90e0348e4ed53b232367d46f92.png)
. Therefore,using again the Stolz-Cesaro theorem we obtain the required limit :
.PP4. Prove that
![$\lim_{n\to\infty}\ \left[\frac {\left(1+\frac 1n\right)^n}{e}\right]^n=\frac {1}{\sqrt e}$](http://latex.artofproblemsolving.com/0/4/b/04b6a8d525ac52448febf3fb6b4b5a950e5947e9.png)
.Proof.
![$l\equiv \lim_{n\to\infty}\ \left[\frac {\left(1+\frac 1n\right)^n}{e}\right]^n=e^L$](http://latex.artofproblemsolving.com/a/4/1/a4176daeab00b95c1c9cbe5f484340f89c06818e.png)
, where ![$L=\frac 1e\cdot \lim_{n\to\infty}\ n\left[\left(1+\frac 1n\right)^n-e\right]\ \stackrel{(*)}{=}\ \frac 1e\cdot\left(-\frac e2\right)=-\frac 12$](http://latex.artofproblemsolving.com/1/7/e/17e28a39a99d51f8fa7f743a719728d0f5d20837.png)
. In conclusion, 
.=====================================================================================================================
![$\blacktriangleright\ (*)\ \lim_{n\to\infty}\ n\cdot\left[\left(1+\frac 1n\right)^n-\text{e}\right]=$](http://latex.artofproblemsolving.com/a/7/7/a779b24fd207dce454f8c379fc3d9d89e46ba570.png)
![$\lim_{n\to\infty}\ n\text{e}\cdot\left[\frac 1{\text{e}}\left(1+\frac 1n\right)^n-1\right]=$](http://latex.artofproblemsolving.com/9/3/a/93a2b796b8b378e64a713974537d859ada876325.png)
![$\text{e}\cdot\lim_{n\to\infty}\ n\cdot\left[\text{e}^{n\ln\left(1+\frac 1n\right)-1}-1\right]=$](http://latex.artofproblemsolving.com/5/2/3/523b68a4120f509c795a9531f14af311f97baabb.png)
![$\text{e}\cdot\lim_{n\to\infty}\ n\cdot\left[n\ln\left(1+\frac 1n\right)-1\right]=$](http://latex.artofproblemsolving.com/8/1/f/81f53ee73d46fa50bd7bd5c45c5351ce2f5cd212.png)
![$-\text{e}\cdot\lim_{n\to\infty}\ \left[n-n^2\ln\left(1+\frac 1n\right)\right]=$](http://latex.artofproblemsolving.com/a/e/a/aeaf7394f7caeb8171b0b2ac2e5c0e11625fe168.png)
, where 
and 
.Proof.
![$\lim_{x\searrow 0}\left[\frac {(1+x)^{\frac 1x}}{e}\right]^{\frac 1x}=e^l$](http://latex.artofproblemsolving.com/5/c/2/5c261d00e36d293b1a07d406d3c8f3ccca9d65df.png)
, where ![$l=\lim_{x\searrow 0}\frac 1x\cdot \left[\frac {(1+x)^{\frac 1x}}{e}-1\right]=$](http://latex.artofproblemsolving.com/6/9/a/69a0122806edfc89bc1c5f7fbc1fb6f313f6cdb6.png)
![$\boxed{\ \lim_{x\searrow 0}\left[\frac {(1+x)^{\frac 1x}}{e}\right]^{\frac 1x}=\frac {1}{\sqrt e}\ }$](http://latex.artofproblemsolving.com/e/9/8/e98a82a6f9bb7021a83bbd4274497d0a3e27629e.png)
.Remark. I"used the remarkable limits
and 
.PP5. Prove that
, where 
.Proof. Prove easily that
, where 
and 
. Using the well-known limits 
(Euler's number) and
obtaim that 
.Remark. Ascertain analogously that
.PP6. Let
and 
, for any 
. Prove that 
.Proof. The recurrence relation can be written as
and 
. Observe that 
, so 
.But
. From Stolz-Cesaro theorem obtain that : 
.
PP7.
![${\color{white}{.}}\implies\lim_{n\to\infty}\, \left(\sqrt[n+1]{\frac {x_{n+1}}{y_{n+1}}}-\sqrt[n]{\frac {x_n}{y_n}}\ \right)=\frac a{b\cdot \text{e}}$](http://latex.artofproblemsolving.com/0/9/8/098820d616b6d6d937025fc6d0620071d272a738.png)
Remark. In particular, for
and 
we obtain Traian Lalescu's sequence i.e. ![$\lim_{n\to\infty}\, \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\frac 1{\text{e}}$](http://latex.artofproblemsolving.com/2/3/3/233f1a07338a0b165c74280c32cb813a8df4e177.png)
Proof. Let us denote :
![$a_n=\sqrt[n]{\frac {x_n}{y_n}}$](http://latex.artofproblemsolving.com/6/a/a/6aa5b61299c7836dad2ffeba7c01d65b8576d53b.png)
. One can easily show, by Cauchy-d'Alembert criterion that : 
. Moreover, we can obtain that :![$\left\{\ \begin{array}{ccc}
\lim\limits_{n\to\infty}\, \frac {a_{n+1}}{a_n}=\lim\limits_{n\to\infty}\, \left[\frac {a_{n+1}}{n+1}\cdot\frac n{a_n}\cdot\frac {n+1}n\right]=\frac a{b\text{e}}\cdot\frac {b\text{e}}a\cdot 1=1 \\ \\
\lim\limits_{n\to\infty}\, \left(\frac {a_{n+1}}{a_n}\right)^n=\lim\limits_{n\to\infty}\, \left(\frac {x_{n+1}}{n^2x_n}\cdot\frac {ny_n}{y_{n+1}}\cdot n\sqrt[n]{\frac {y_n}{x_n}}\right)^{\frac n{n+1}}=\frac ab\cdot\frac {b\text{e}}a=\text{e}\ \end{array}\right\|$](http://latex.artofproblemsolving.com/c/3/d/c3d0161b9a4128fdb4840f990d1252185668a22f.png)
. On the other hand, note that : ![$ \lim_{n\to\infty}\, \left(\sqrt[n+1]{\frac {x_{n+1}}{y_{n+1}}}-\sqrt[n]{\frac {x_n}{y_n}}\ \right)=$](http://latex.artofproblemsolving.com/8/e/3/8e32be963a26b86315bf25a4ed9a652232cbc4bc.png)
![$\lim_{n\to\infty}\, \left(a_{n+1}-a_n\right)=\lim_{n\to\infty}\, \left[a_n\cdot\left(\frac {a_{n+1}}{a_n}-1\right)\right]=$](http://latex.artofproblemsolving.com/c/c/4/cc4bf7d706a0f42f16f412338f09c5a227dec943.png)
![$\lim_{n\to\infty}\ \left[\frac {a_n}n\cdot\frac {\text{e}^{\ln\left(\frac {a_{n+1}}{a_n}\right)}-1}{\ln\left(\frac {a_{n+1}}{a_n}\right)}\cdot\ln\left(\frac {a_{n+1}}{a_n}\right)^n\right]=\frac a{b\cdot\text{e}}\cdot 1\cdot\ln\text{e}=\frac a{b\cdot\text{e}}$](http://latex.artofproblemsolving.com/b/d/b/bdbaac9a094819832defe12e9bd66ff2843c1bf1.png)
, as required.PP8. Prove that
and 
, where 
.Proof. One can easily prove by induction that the sequence
has positive terms. On the other hand, by Bernoulli's inequality one gets: 
- note that 
- so the given sequence it is decreasing, thus convergent and let 
be its limit. By passing to limit in the recurrentrelation we obtain:
and differentiating with respect to 
both sides of the latter equality we now have: 
whence the desiredlimit :
. For the second part of our problem let's note that the required limit, say 
, can be written as : 
so this is an indeterminationcase
- because is monotonously decreasing and its limit is equal to 
then 
is monotonously increasing to 
. Now we need to find thelimit
and in case this is finite then, by Stolz-Cesaro theorem, we will conclude that 
. Observe that : 
![$\lim_{n\to\infty}\, \frac {x_n\left[\left(1+\frac {x_n}a\right)^a-1\right]}{x_n-\left(1+\frac {x_n}a\right)^a+1}=$](http://latex.artofproblemsolving.com/d/d/e/dde0d4ea9703a5043ab058e6184725c597d19365.png)
![$\lim_{n\to\infty}\left[ \frac {\left( 1+\frac {x_n}a\right )^{a-1}}{\frac {x_n}a\cdot\frac {x_n^2}a\cdot\frac 1{1+x_n}-\left(1+\frac {x_n}a\right)^a} \right]=$](http://latex.artofproblemsolving.com/9/d/e/9def6b1eedfb72fb1b5ba40fccd03efccf7f1834.png)
. But
so
and accordingly 
.PP9. Is well-known that
, where 
. Prove that 
and 
.Proof.
, where 
. 
, where 
. PP10. Prove that
, where 
for 
.Proof.
,where
and 
. In conclusion, 
and 
.PP11. Prove that
![$\lim_{n\to\infty}\sqrt [n]{\frac {C_{3n}^n}{C_{2n}^n}}=$](http://latex.artofproblemsolving.com/e/8/2/e82e41a123fd955a880f6fb1e38f08bca9dd0b9f.png)
. In general, ![$\{p,q\}\subset\mathbb N\ ,\ \ p>q\ge 2\ \implies\ \lim_{n\to\infty}\ \sqrt [n]{\frac {C_{pn}^n}{C_{qn}^n}}=\frac {p^p(q-1)^{q-1}}{q^q(p-1)^{p-1}}$](http://latex.artofproblemsolving.com/6/8/2/682fa9d40c1cb1b3cfa98c39ac35759dd9163050.png)
.Proof.
![$a_n=\sqrt [n]{x_n}$](http://latex.artofproblemsolving.com/d/b/b/dbb2c2b54872b62a5d0f7a99e4f2361f4518f30a.png)
, 
. Since 
, from an well-known property obtain that and ![$\boxed{a_n=\sqrt [n]{x_n}\rightarrow \frac {27}{16}}$](http://latex.artofproblemsolving.com/6/d/8/6d804daba89e491919b538c8bac8921d22063fac.png)
.PP12. Find
![$\lim_{n\to\infty} n^{b-a}\cdot\sqrt[n]{\frac{(an)!}{(bn)!}}$](http://latex.artofproblemsolving.com/c/8/6/c86a833c5211318cd03746962e95463076b56249.png)
, where 
.Proof. If
![$a_n=\sqrt[n]{b_n}$](http://latex.artofproblemsolving.com/8/a/7/8a7c7e1a52ba6edbce28b8432a61c94fdf475065.png)
, where 
, then 
and
. In conclusion, using the well-known property ![$\frac {b_{n+1}}{b_n}\rightarrow l\implies \sqrt [n]{b_n}\rightarrow l$](http://latex.artofproblemsolving.com/c/7/c/c7c1eaa4265d234bfad345eb41eeb102e856a915.png)
, obtain that 
.PP13. Prove that
![$\boxed{\, L=\lim_{x\to 0}\, \frac {\left[\arcsin\left(\frac {\sin x}x\right)-\arcsin\left(1\right)\right]^2}{\arctan\left(\frac {\tan x}x\right)-\arctan\left(1\right)}=2\, }$](http://latex.artofproblemsolving.com/9/2/a/92aaaa954ee9b1119c9ecbfad4d22a5d4c619ad0.png)
.Proof.
as well as: 
. Thus,![$L=\lim_{x\to 0}\, \left[\left(\frac {-\arcsin\left(\sqrt {1-\left(\frac {\sin x}x\right)^2}\right)}{\sqrt{1-\left(\frac {\sin x}x\right)^2}}\right)^2\cdot\frac {\frac {\tan x-x}{\tan x+x}}{\arctan\left(\frac {\tan x-x}{\tan x+x}\right)}\cdot\frac {1-\left(\frac {\sin x}x\right)^2}{\frac {\tan x-x}{\tan x+x}}\right]$](http://latex.artofproblemsolving.com/7/f/a/7fa4e4d82887f5cc3090fe8e5314857329779044.png)
whence, by taking into account that
, our limit reduces to computing 
.On the other hand, one can easily prove that
. Now it is easy to see that the desired limit can be conveniently written as![$L=\lim_{x\to 0}\, \left[\frac {x-\sin x}{x^3}\cdot\frac {x^3}{\tan x-x}\cdot\frac {x+\sin x}x\cdot\frac {x+\tan x}x\right]$](http://latex.artofproblemsolving.com/1/f/2/1f2e638dacd2a3278140c3ca43e37b00d40ff149.png)
which finally equals 
because
. Here is another related limit, which is equally interesting ![$\boxed{\, L_1=\lim_{x\to 0}\, \frac {\left[\arcsin\left(\frac {\tan x}x\right)-\arcsin\left(1\right)\right]^2}{\arctan\left(\frac {\sin x}x\right)-\arctan\left(1\right)}=8\, }$](http://latex.artofproblemsolving.com/b/8/d/b8d6c6f7ec573a34fead37c6dc212931cd332c8b.png)
.PP14. Calculate the limit
![$\lim_{x\to\infty}x\left[\left(\frac {x}{x+1}\right)^x-\frac 1e\right]$](http://latex.artofproblemsolving.com/7/6/1/761373ecaf4d7c5ec1af26cde3ec66143b875a6a.png)
.Proof.
![$\lim_{x\to\infty}x\left[\left(\frac {x}{x+1}\right)^x-\frac 1e\right]=$](http://latex.artofproblemsolving.com/7/4/8/74885a576e3a02fa8525554bf013568b7546fc0a.png)
![$\frac 1e\cdot\lim_{x\to\infty}x\left[e\left(\frac {x}{x+1}\right)^x-1\right]\stackrel{(*)}{=}$](http://latex.artofproblemsolving.com/b/d/9/bd9c2621430cb3fde67518efea9360cfcbbe9b58.png)
![$\frac 1e\cdot\lim_{x\to\infty}x\ln\left[e\left( \frac {x}{x+1}\right)^x\right]=$](http://latex.artofproblemsolving.com/5/d/f/5dfa2250a9f5be78a3a00918f1c7e894662cee57.png)
![$\frac 1e\cdot\lim_{x\to\infty}x^2\left[\frac 1x+\ln x- \ln (x+1)\right]=$](http://latex.artofproblemsolving.com/5/1/2/512211e532a0d8f5e15c57dd5cbc957ef65ce955.png)
.Rematk. In upper equality
I used the remarkable limits 
and 
.PP15. For
find ![$\lim_{x\to \infty}\left[\left(x-x^p\right)^p-x^p\right]$](http://latex.artofproblemsolving.com/4/e/3/4e32df89287cf9ef6d84776f07e7f9b841132f6e.png)
.Proof.
![$\lim_{x\to\infty}\left[\left(x-x^p\right)^p-x^p\right]=$](http://latex.artofproblemsolving.com/3/2/6/326a66d70ad1572ea30e96ce2c3be5b4ac860894.png)
![$\lim_{x\to\infty}x^p\left[\left(1-x^{p-1}\right)^p-1\right]\stackrel{(*)}{=}$](http://latex.artofproblemsolving.com/0/4/a/04a8dc0584e2a052fa2b3707f52f984e5a19c17f.png)
.Remark. In the equality
I used the remarkable limit 
.PP16. Ascertain the limit
![$l=\lim_{n\to\infty}\left[\prod_{k=1}^n\left(1+\frac {2k-1}{2n}\right)\right]^{\frac {1}{2n}}$](http://latex.artofproblemsolving.com/8/1/d/81dd8359f6e7373f537b11fcca410e74e91cc1c9.png)
.Proof.
![$\ln 2-\left|\left[x-\ln (x+1)\right]\right|_0^1=$](http://latex.artofproblemsolving.com/d/6/7/d67946b140175de16827dbed6adbeb4e7cddbcff.png)
.PP17 (without involving l'Hospital rules) Ascertain
![$L=\lim_{x\to\infty} x\cdot \left(\sqrt[p]{x^p+ax^{p-1}+bx^{p-2}+\ \ldots}+\sqrt[p]{x^p-ax^{p-1}+cx^{p-2}+\ \ldots}-2x\right)$](http://latex.artofproblemsolving.com/a/9/0/a908509bd9d667f2cdc6bbb09a3a97963795f087.png)
.Proof. Denote
![$\left\{\begin{array}{c}
f(x)=\sqrt[p]{x^p+ax^{p-1}+bx^{p-2}+\ \ldots}\\\\
g(x)=\sqrt[p]{x^p-ax^{p-1}+cx^{p-2}+\ \ldots}\end{array}\right\|$](http://latex.artofproblemsolving.com/a/1/1/a11062e9f6dedd5627beaa289a42abaf5fc987e8.png)
. Prove easlily that ![$\lim_{x\to\infty}\ \left[f(x)-x\right]=\frac ap$](http://latex.artofproblemsolving.com/7/a/6/7a6a7110e9709b58186979a4dc5016ab13d294ff.png)
and ![$\lim_{x\to\infty}\ \left[g(x)-x\right]=-\frac ap$](http://latex.artofproblemsolving.com/6/a/c/6ac7a358fd1d9a5ecbf78951e6ab336f115b0810.png)
. Hence we are in the exception case 
. Using the substitution 
obtain that ![$L=\lim_{y\to 0}\frac {1}{y^2}\cdot \left(\sqrt[p]{1+ay+by^2+\ \ldots}+\sqrt[p]{1-ay+cy^2+\ \ldots}-2\right)$](http://latex.artofproblemsolving.com/f/e/e/fee7907bdb591f018f8576c347a6d837d8622453.png)
. Denote ![$\left\{\begin{array}{c}
u(y)=\sqrt [p]{1+ay+by^2+\ \ldots}\\\\
v(y)=\sqrt [p]{1-ay+cy^2+\ \ldots}\end{array}\right\|$](http://latex.artofproblemsolving.com/1/3/c/13caaab4be1b2431ea5cccd4acba7fcc81e45989.png)
. Observe that 
and 
. Since ![$u(y)+v(y)-2=u(y)v(y)-1-[u(y)-1][v(y)-1]$](http://latex.artofproblemsolving.com/7/d/f/7df196b3a725de4ed97c12010f0515f35700349f.png)
obtain that ![$L=\lim_{y\to 0}\ \left[\frac {u(y)v(y)-1}{y^2}-\frac {u(y)-1}{y}\cdot\frac {v(y)-1}{y}\right]=\frac 1p\cdot \left(b+c-a^2\right)+\frac {a^2}{p^2}$](http://latex.artofproblemsolving.com/9/b/d/9bd4be9cbc68e124cc1e01ef715f8ce505f7c43f.png)
. In conlusion, 
.Remark. In the particular case obtain that
![$\lim_{x\to\infty}\ x\cdot\left(2x-\sqrt [3]{x^3+x^2+1}-\sqrt [3]{x^3-x^2+1}\right)=\frac 29$](http://latex.artofproblemsolving.com/2/d/a/2daf8d4f5897e0adceb4f622e868f859a73e7a84.png)
.PP18 Prove that
![$\lim_{n\to\infty}\ \frac {n\left(\sqrt[n]n-1\right)}{\ln n}=$](http://latex.artofproblemsolving.com/4/1/b/41bee6d7db7d19924d3aeb9d9d99c4f6f4a99380.png)
![$1\ ,\ \lim_{n\to\infty}\ n\Big(\sqrt[n]{n}-1 \Big)-\ln{n}$](http://latex.artofproblemsolving.com/5/0/a/50a63c0d3e0a7ba95b37658a319b427f12b1d694.png)
![$=0\ ,\ \boxed{\lim_{n\to\infty}\ \frac {n}{\ln^2n}\cdot \Big[n\Big(\sqrt[n]{n}-1 \Big)-\ln{n}\Big]=\frac 12}$](http://latex.artofproblemsolving.com/d/5/1/d5153161abd8a05418b385b7dfbe3d4725418ba2.png)
.Proof. Since
, for 
obtain that 
![$\lim_{n\to\infty}\ \frac {n^2}{\ln^2n}\cdot\left(\sqrt[n]n-1-\frac {\ln n}{n}\right)=\frac 12\implies $](http://latex.artofproblemsolving.com/f/e/c/fec43978e79db7cb300ed26ad0502a1d421b17c2.png)
![$\lim_{n\to\infty}\ \frac {n}{\ln^2n}\cdot \Big[n\Big(\sqrt[n]{n}-1 \Big)-\ln{n}\Big]=\frac 12$](http://latex.artofproblemsolving.com/5/3/1/531be8aa94cee66a7e2bce1f2b8cc265205f0568.png)
.PP19. Let
be a sequence defined by 
and 
for any . Prove that 
and find 
.Proof. From
we have 
. Squaring gives 
or 
with 
and 
. For (a) it suffices to show that 
for all
. This follows immediately from 
and 
. For (b) it suffices to show that 
. Because 
and from (a) , we have
. It is well-known that the harmonic series 
is less than
for some constant 
(Approximate the series by 
to see this). As a result, 
is bounded between 
where 
. Now since
, it is clear that 
which implies the desired result.PP20. Prove that
, where 
.Proof. Using the inequality
obtain that 
. Therefore,
.See here
This post has been edited 119 times. Last edited by Virgil Nicula, Jul 17, 2017, 8:42 AM
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