22. Some interesting limits (sequence/function).
by Virgil Nicula, Apr 22, 2010, 5:04 PM
0I. Exercises with limits (sequence/function).
. Number 
. Define 
, where 
.
, i.e. ![$a_n\nearrow a\in \overline R \equiv [-\infty ,\infty ].$](//latex.artofproblemsolving.com/e/7/b/e7b4978a109fcce9f52404569c2734d8ef83c54e.png)
Apply the A.M.-G.M. to 
and 
, i.e. 
Apply the A.M.-G.M. to 
. Apply Weierstrass' theorem: 
:
. Thus, 
.
and 
.
Applications. 
(Euler's number) 
.
![$\left(\sqrt [p]{1+\frac{1}{n}}+\sqrt [p]{1-\frac{1}{n}}-2\right)=-\frac{p-1}{p^{2}}$](//latex.artofproblemsolving.com/d/3/7/d3705d36991ffdd4d8ca4afeb14b00a8386ab515.png)
.
Generally (without l'Hospital's rules) : 
![$\lim_{x\to 0}\frac{1}{x^{2}}\cdot\left(\sum_{k=1}^{n}\sqrt [p]{1+a_{k}x+b_{k}x^{2}+\ldots }-n\right)=\frac{1}{p}\cdot\sum_{k=1}^{n}b_{k}-\frac{p-1}{2p^{2}}\cdot\sum_{k=1}^{n}a^{2}_{k}$](//latex.artofproblemsolving.com/0/5/5/05558124ff0c336510538fb56e2dca9a1c2aa7a8.png)
.
(Isac Schoenberg).
Proof. Prove easily that
and 
. Particularly, for any 
, 
, i.e. 
. Therefore,
. For a fixed 
we have
For 
obtain 
. For 
obtain 
.
From the relations
and 
obtain 
.
![$ \frac {1}{\sqrt [4]e}$](//latex.artofproblemsolving.com/d/3/9/d39b0bc8135c92009ce402fc5c736214e15858b0.png)
.
Proof. Define
. Observe that 
and we are in the indetermination 
. Therefore, 
, where
, where 
. Define
and 
. Observe that 
and ![$ c_n = \frac {1}{n\sqrt {n^2 + n} + n^2 + n}\cdot\sum_{k = 1}^nk = \frac {n(n + 1)}{2\left[n\sqrt {n^2 + n} + n^2 + n\right]}\rightarrow\frac 14$](//latex.artofproblemsolving.com/1/1/6/116382a473b0592c80333d3cac080ec45ad59d25.png)
and![$ d_n = \frac {1}{n\sqrt {n^2 + 1} + n^2 + 1}\cdot\sum_{k = 1}^nk = \frac {n(n + 1)}{2\left[n\sqrt {n^2 + 1} + n^2 + 1\right]}\rightarrow\frac 14$](//latex.artofproblemsolving.com/0/e/1/0e16fbc4f5a4fe0afd1543c1fa549088b55f9377.png)
. In conclusion, 
and 
, i.e. ![$ L = \frac {1}{\sqrt [4]{e}}$](//latex.artofproblemsolving.com/9/f/7/9f77bcd074210c5f10f6e054f1f5ca18114a9aeb.png)
.
![$\lim_{n\to\infty}a_n=\frac {1}{\sqrt [4]e}$](//latex.artofproblemsolving.com/e/0/0/e007bb0136ca8d96eac56cf27bd9ddd231b67599.png)
, where 
.
Proof. Observe that![$a_n=\left[\prod_{k=1}^n\left(\frac kn\right)^{\frac kn}\right]^{\frac 1n}$](//latex.artofproblemsolving.com/5/8/8/5880ae1937a6d1efc9d7c41567bf0596345368ec.png)
and 
. In conclusion, ![$a_n\rightarrow\frac {1}{\sqrt [4]e}$](//latex.artofproblemsolving.com/1/8/3/1833a4899d65211102f5fd5b99b54f3bb9a63279.png)
.
(Mateescu Constantin). ![$\lim_{n\to\infty}\ \frac {\sqrt{1+\frac 1{\ln 2}}+\sqrt[3]{1+\frac 1{\ln 3}}+\ldots +\sqrt[n]{1+\frac 1{\ln n}}-n}{\ln (\ln n)}=1$](//latex.artofproblemsolving.com/4/8/6/486af4928b81a0617edf698d7460f3be74eb561e.png)
.
Proof.![$\lim_{n\to\infty}\frac {1}{\ln(\ln n)}\cdot\left[\sum_{k=2}^{n}\sqrt[k]{1+\frac{1}{\ln k}-n}\right]\ \stackrel{\mathrm{S.C.}}{=}$](//latex.artofproblemsolving.com/c/6/0/c607f587a73cefb444cabc38fb441414010062f7.png)
![$\lim_{n\to\infty}\frac{\sqrt[n+1]{1+\frac{1}{\ln (n+1)}}-1}{\ln\frac{\ln (n+1)}{\ln n}}=$](//latex.artofproblemsolving.com/a/e/0/ae088b1b846f0b4baf8e5ba4b7d18b50ff1e6b28.png)
![$\lim_{n\to\infty}\frac{e^{\frac{1}{n+1}\cdot\ln\left [1+\frac{1}{\ln (n+1)}\right]}-1}{\ln\frac{\ln (n+1)}{\ln n}}=$](//latex.artofproblemsolving.com/e/4/b/e4be34c28f8d04f977c34a6d4eddb0444a41d782.png)
![$\lim_{n\to\infty}\frac{\frac{1}{n+1}\cdot\ln\left [1+\frac{1}{\ln (n+1)}\right]}{\ln\frac{\ln (n+1)}{\ln n}}=$](//latex.artofproblemsolving.com/3/f/b/3fb2823c7d97a7ce5e226b885e810c28998f1241.png)
![$\lim_{n\to\infty}\frac{\frac{1}{n+1}\cdot\frac{1}{\ln (n+1)}}{\ln\left [1+\frac{\ln (n+1)-\ln n}{\ln n}\right]}
=$](//latex.artofproblemsolving.com/d/7/3/d732af165ba65d945f00ce68201f74b66cc4f730.png)
.
Ascertain 
, where ![$I=\left(0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/a/8/a/a8abc00f9ba0fb210e7d7ee466112d4d2d4d4871.png)
and the correspondence is 
.
Proof. I will find studyng monotony of ![$f(x)=\frac{\sin x-x\cos x}{x-\sin x},x\in \left(0,\frac{\pi}{2}\right]$](//latex.artofproblemsolving.com/2/f/b/2fbfeb2a4946a032b40bb8d1e457e864f0db52f5.png)
. Thus 
![$\left[x-(1-\cos x)\cdot \frac{1+\sqrt{5+4\cos x}}{2\sin x}\right] \equiv\ g(x)$](//latex.artofproblemsolving.com/2/6/9/2699852ba5c6dd63a42f0c7645937338eb3e48c5.png)
. So 
and 
. Thus, 
and 
![$\boxed {\ x\in \left(\ 0\ ,\ \frac {\pi}{2}\ \right]\ \implies\ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2\ \implies\ \mathrm{Im}(f)=\left[\frac {2}{\pi -2},2\right)\ }$](//latex.artofproblemsolving.com/5/9/d/59d6b119ef5c7d4f16bdb88d0b874c752be7ed7b.png)
.
Geometrical interpretation. Let
such that 
and 
for which 
and 
. For 
, where 
and 
, let 
so that
and 
. I"ll seek position of the remotest (to left) 
(if it exists) such that for any , have the length 
. Thus, 
, i.e. 
. Therefore, the length 
for any ![$x\in\left(0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/c/b/9/cb9b8254c4dd909eac5d5dc6affd8c6ab5c35f69.png)
. With other words, the left remotest 
for which
![$\left(\forall\right) x\in\left(0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/a/8/a/a8af00d00d87ceb0b67dc9dafff2625d4115ff78.png)
we have the length of 
is 
, i.e. 
because 
is decresing and 
and 
.
Applications. Prove that the following inequalities :![$x\in \left(0,\frac{\pi}{2}\right]\Longrightarrow\frac{2}{\pi}\le\boxed {\left(1-\frac{2}{\pi}\right)\cos x+\frac{2}{\pi}\le \frac{\sin x}{x}<\frac{2+\cos x}{3}}\le 1\ ;$](//latex.artofproblemsolving.com/7/f/1/7f106e00a7ec3db546f52d6bce000625d5aa373f.png)
.
Remark. I used the equivalence relation ("same signature") over
Compute 
, where 
.
Proof. The given limit can be written as
, where 
.
. I used the remarkable limits 
, 
and 
.
.
I used the remarkable limits and 
. In conclusion, 
.
Proposed problem. Find 
so that 
.
Proof. I"ll use only the remarkable limit
, where 
. Observe that 
. In conclusion, our limit
and in this case 
.
PP. 
![$L_1=\lim_{x\to 0}\, \frac {\text{e}^{\frac {\ln (1+ax)}x}-\text{e}^a}x=\lim_{x\to 0}\, \left[\text{e}^a\cdot\frac {\text{e}^{\frac {\ln (1+ax)}x-a}-1}{\frac {\ln (1+ax)}x-a}\cdot\frac {\ln\left(1+ax\right)-ax}{x^2}\right]\stackrel{\text{l'H}}{=}$](//latex.artofproblemsolving.com/7/8/4/784ad9c2d1a92a3cf4d26a97c4c2ccbb78492329.png)
![$L_2=\lim_{x\to 0}\, \frac {\text{e}^{\frac {a\ln (1+x)}x}-\text{e}^a}x=\lim_{x\to 0}\, \left[\text{e}^a\cdot\frac {\text{e}^{\frac {a\ln (1+x)}x-a}-1}{\frac {a\ln (1+x)}x-a}\cdot\frac {a\ln\left(1+x\right)-ax}{x^2}\right]\stackrel{\text{l'H}}{=}$](//latex.artofproblemsolving.com/e/d/a/eda3d441c2d40e5e7937da16783d45b9f1114921.png)
.
PP. Ascertain 
, where 
.
Proof.
.
I used the remarkable limits
and 
.
PP. Ascertain 
, where 
.
Proof.
, where 
.
PP. Solve recurrence system 
, , where 
.
Proof. Define for any the matrix 
, where 
and 
where 
.
Observe that
for any and 
, where 
and 
.
Generally, for any we have 
, where and 
. Therefore,
the associated equation
has the roots 
. Thus, for any we have
and 
and we can ascertain easily these four constans 
,
where
from the first two terms of each from these two sequencies.
PP. Using eventually the well-known limit 
prove that
, where ![$a_n=\frac {\sqrt[n]{n!}}{n}$](//latex.artofproblemsolving.com/0/6/8/06820a0dbd99d51537b0e036ac49ce0431c59ba8.png)
. Deduct that ![$\lim_{n\to\infty}\frac {n}{\ln n}\cdot\left(e-\frac {n}{\sqrt [n]{n!}}\right)=\frac e2$](//latex.artofproblemsolving.com/c/e/f/cef8973b9bfa541238d38c06fbd3d47ab2bd9735.png)
.
Proof.
, where 
,
and 
. Therefore, 
. Thus, ![$\lim_{n\to\infty}\frac {n}{\ln n}\cdot\left(e-\frac {n}{\sqrt[n]{n!}}\right)=$](//latex.artofproblemsolving.com/3/e/2/3e2fa090577cb1506572edf4c32f7ccba08a9627.png)
![$\lim_{n\to\infty}\frac {n}{\ln n}\cdot\left(\frac {\sqrt[n]{n!}}{n}-\frac 1e\right)\cdot\frac {ne}{e\sqrt[n]{n!}-n}\cdot\frac {e\sqrt[n]{n!}-n}{\sqrt[n]{n!}}=$](//latex.artofproblemsolving.com/3/e/e/3eefba40682899fb393acba9880b63bd09879d7b.png)
![$\frac {1}{2e}\cdot\lim_{n\to\infty}\frac {ne}{\sqrt[n]{n!}}=\frac e2$](//latex.artofproblemsolving.com/a/c/a/aca4ba31c8fc08ebe1a54c37c228b0a4198b61d8.png)
.
PP. Let 
be the root of 
, where 
, 
. Prove that 
.
Proof. Prove easilky that
for any and 
, i.e. ![$x_n > \sqrt[n]{p+1}$](//latex.artofproblemsolving.com/c/f/5/cf5af3d30ad4d0cb5e5ebfec33297459a47e4dc9.png)
. We can set similarly 
,
i.e.![$x_n < \sqrt[n-p]{p+1}$](//latex.artofproblemsolving.com/9/5/7/9573bdca2f3eba7629896b493a62525f35a305a6.png)
. Thus, ![$n\left(\sqrt[n]{p+1} - 1\right) < n\left(x_n - 1\right) < n\left(\sqrt[n-p]{p+1} - 1\right)$](//latex.artofproblemsolving.com/1/2/5/125de158179c9148728390edbb94a5d3f39c6656.png)
and from the squeeze theorem, the limit is 
.
PP. Evaluate ![$\lim_{n\to \infty} \frac{n^2}{\ln n}\cdot \left(\sqrt[n]{n}-\sqrt[n+1]{n+1}\right)$](//latex.artofproblemsolving.com/9/3/a/93acf477a8c4bbc8f99955f80f7666d0092b757f.png)
.
Proof.![$\lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot \left(\sqrt[n]{n}-\sqrt[n+1]{n+1}\right)=$](//latex.artofproblemsolving.com/5/5/4/55408e413e08851f020bc898a2e4d3588a6d93ed.png)
![$\lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot \sqrt [n+1]{n+1}\cdot\left(\frac {\sqrt[n]n}{\sqrt[n+1]{n+1}}-1\right)=$](//latex.artofproblemsolving.com/f/5/3/f53ee98a924d6b5a676ef2347d55701a8c4e092f.png)
![$\lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot\frac {\frac {\sqrt[n]n}{\sqrt[n+1]{n+1}}-1}{\ln\frac {\sqrt[n]n}{\sqrt[n+1]{n+1}}}\cdot \left[\frac 1n\cdot\ln n-\frac {1}{n+1}\cdot\ln (n+1)\right]=$](//latex.artofproblemsolving.com/c/2/e/c2e4800b78e5d9feea5ab601dc205fe881fcd127.png)
.
In conclusion,![$\boxed{\ \lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot \left(\sqrt[n]{n}-\sqrt[n+1]{n+1}\right)=1\ }$](//latex.artofproblemsolving.com/6/9/b/69b14821da6de474bd1686cc2fd5e415b6fc223e.png)
. I used the remarkable limits ![$\lim_{n\to\infty}\ \sqrt [n]n=1$](//latex.artofproblemsolving.com/3/6/d/36de8bf895ee5ffcabd891430c63b04c17347b0b.png)
, 
and 
.
Calculate ![$\lim_{x\to\infty}\ x\cdot\left[\left(\frac {a^{\frac 1x}+b^{\frac 1x}}{2}\right)^x-\sqrt{ab}\right]$](//latex.artofproblemsolving.com/0/a/8/0a8a932ceef259e3cfc8297c29917c79f70cd48a.png)
, where 
.
Proof. Observe that
, where 
.
![$\lim_{x\to\infty}\ x\cdot\left[\left(\frac {a^{\frac 1x}+b^{\frac 1x}}{2}\right)^x-\sqrt{ab}\right]\stackrel{(\infty\cdot 0)}{\ =\ }$](//latex.artofproblemsolving.com/4/b/0/4b0552baebb528ed231bb54bfc812857af079822.png)
![$\lim_{x\to 0}\ \frac 1x\cdot\left[\left(\frac {a^x+b^x}{2}\right)^{\frac 1x}-\sqrt{ab}\right]\stackrel{(\frac 00)}{\ =\ }$](//latex.artofproblemsolving.com/9/6/1/961cfbdd2aab59b6c7e8975ca2cfbad4cc4cc90e.png)
![$\sqrt{ab}\cdot \lim_{x\to 0}\ \frac 1x\cdot\left[\frac 1x\cdot\ln\frac {a^x+b^x}{2}-\frac 12\cdot\ln (ab)\right]=$](//latex.artofproblemsolving.com/a/0/4/a0484d9899cb366ac17e01f9e1d6414366155597.png)
![$\frac {\sqrt{ab}}{2}\cdot\lim_{x\to 0}\ \frac {1}{x^2}\cdot\left[2\cdot\ln\frac {a^x+b^x}{2}-x\ln (ab)\right]\stackrel{(l'H)}{=}$](//latex.artofproblemsolving.com/0/b/1/0b1c351ffeaf7d7170be6a4dfb0dc5c4a2815068.png)
![$\frac {\sqrt{ab}}{2}\cdot\lim_{x\to 0}\ \frac {1}{2x}\cdot\left[\frac {2}{a^x+b^x}\cdot\left(a^x\cdot\ln a+b^x\cdot\ln b\right)-\ln (ab)\right]=$](//latex.artofproblemsolving.com/c/5/4/c5407a66011a128c49a49f99b5b01632e65c6013.png)
![$\frac {\sqrt{ab}}{8}\cdot\lim_{x\to 0}\ \frac 1x\cdot\left[2\left(a^x\ln a+b^x\ln b\right)-\left(a^x+b^x\right)(\ln a+\ln b)\right]=$](//latex.artofproblemsolving.com/3/9/b/39b9a5449adb875199e1d5543772b483be3088e7.png)
![$\boxed{\ \lim_{x\to\infty}\ x\cdot\left[\left(\frac {a^{\frac 1x}+b^{\frac 1x}}{2}\right)^x-\sqrt{ab}\right]=\frac {\sqrt{ab}}{8}\cdot\ln^2\frac ab\ }$](//latex.artofproblemsolving.com/5/9/9/599033a277a919cf98dc4c14fa408b102a96b0c8.png)
.
Without using L'Hospital's Rules, find 
.
Proof. Denote
. Prove easily that 
. If 
(it exists
and it is finite), then for
in the relation 
obtain that 
, i.e. 
. Therefore, 
. In conclusion, 
.
II. Siruri recurente.
Articol din G.M.B. - o clasa de siruri recurente (<== click).
Examples. Ascertain the limits of the following strings 
, where 
:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
!2.
13.
14.
15.
16.
17.
18.
19.![$x_1>1,\ x_{n+1}=\sqrt [3] {3x_n-2},\ y_n=n(x_n-1)\ .$](//latex.artofproblemsolving.com/3/4/4/344a8ccb25f7ef8fe2046f2120a11c3c3e26974e.png)
20.
See here and here
. Number 
. Define 
, where 
.
, i.e. ![$a_n\nearrow a\in \overline R \equiv [-\infty ,\infty ].$](http://latex.artofproblemsolving.com/e/7/b/e7b4978a109fcce9f52404569c2734d8ef83c54e.png)
Apply the A.M.-G.M. to 
and 
, i.e. 
Apply the A.M.-G.M. to 
. Apply Weierstrass' theorem: 
:
. Thus, 
.
and 
.
Applications. 
(Euler's number) 
.
![$\left(\sqrt [p]{1+\frac{1}{n}}+\sqrt [p]{1-\frac{1}{n}}-2\right)=-\frac{p-1}{p^{2}}$](http://latex.artofproblemsolving.com/d/3/7/d3705d36991ffdd4d8ca4afeb14b00a8386ab515.png)
.
Generally (without l'Hospital's rules) : 
![$\lim_{x\to 0}\frac{1}{x^{2}}\cdot\left(\sum_{k=1}^{n}\sqrt [p]{1+a_{k}x+b_{k}x^{2}+\ldots }-n\right)=\frac{1}{p}\cdot\sum_{k=1}^{n}b_{k}-\frac{p-1}{2p^{2}}\cdot\sum_{k=1}^{n}a^{2}_{k}$](http://latex.artofproblemsolving.com/0/5/5/05558124ff0c336510538fb56e2dca9a1c2aa7a8.png)
.
(Isac Schoenberg).Proof. Prove easily that
and 
. Particularly, for any 
, 
, i.e. 
. Therefore,
. For a fixed 
we have
For 
obtain 
. For 
obtain 
.From the relations
and 
obtain 
.
![$ \frac {1}{\sqrt [4]e}$](http://latex.artofproblemsolving.com/d/3/9/d39b0bc8135c92009ce402fc5c736214e15858b0.png)
.Proof. Define
. Observe that 
and we are in the indetermination 
. Therefore, 
, where
, where 
. Define
and 
. Observe that 
and ![$ c_n = \frac {1}{n\sqrt {n^2 + n} + n^2 + n}\cdot\sum_{k = 1}^nk = \frac {n(n + 1)}{2\left[n\sqrt {n^2 + n} + n^2 + n\right]}\rightarrow\frac 14$](http://latex.artofproblemsolving.com/1/1/6/116382a473b0592c80333d3cac080ec45ad59d25.png)
and
![$ d_n = \frac {1}{n\sqrt {n^2 + 1} + n^2 + 1}\cdot\sum_{k = 1}^nk = \frac {n(n + 1)}{2\left[n\sqrt {n^2 + 1} + n^2 + 1\right]}\rightarrow\frac 14$](http://latex.artofproblemsolving.com/0/e/1/0e16fbc4f5a4fe0afd1543c1fa549088b55f9377.png)
. In conclusion, 
and 
, i.e. ![$ L = \frac {1}{\sqrt [4]{e}}$](http://latex.artofproblemsolving.com/9/f/7/9f77bcd074210c5f10f6e054f1f5ca18114a9aeb.png)
.
![$\lim_{n\to\infty}a_n=\frac {1}{\sqrt [4]e}$](http://latex.artofproblemsolving.com/e/0/0/e007bb0136ca8d96eac56cf27bd9ddd231b67599.png)
, where 
.Proof. Observe that
![$a_n=\left[\prod_{k=1}^n\left(\frac kn\right)^{\frac kn}\right]^{\frac 1n}$](http://latex.artofproblemsolving.com/5/8/8/5880ae1937a6d1efc9d7c41567bf0596345368ec.png)
and 
. In conclusion, ![$a_n\rightarrow\frac {1}{\sqrt [4]e}$](http://latex.artofproblemsolving.com/1/8/3/1833a4899d65211102f5fd5b99b54f3bb9a63279.png)
.
(Mateescu Constantin). ![$\lim_{n\to\infty}\ \frac {\sqrt{1+\frac 1{\ln 2}}+\sqrt[3]{1+\frac 1{\ln 3}}+\ldots +\sqrt[n]{1+\frac 1{\ln n}}-n}{\ln (\ln n)}=1$](http://latex.artofproblemsolving.com/4/8/6/486af4928b81a0617edf698d7460f3be74eb561e.png)
.Proof.
![$\lim_{n\to\infty}\frac {1}{\ln(\ln n)}\cdot\left[\sum_{k=2}^{n}\sqrt[k]{1+\frac{1}{\ln k}-n}\right]\ \stackrel{\mathrm{S.C.}}{=}$](http://latex.artofproblemsolving.com/c/6/0/c607f587a73cefb444cabc38fb441414010062f7.png)
![$\lim_{n\to\infty}\frac{\sqrt[n+1]{1+\frac{1}{\ln (n+1)}}-1}{\ln\frac{\ln (n+1)}{\ln n}}=$](http://latex.artofproblemsolving.com/a/e/0/ae088b1b846f0b4baf8e5ba4b7d18b50ff1e6b28.png)
![$\lim_{n\to\infty}\frac{e^{\frac{1}{n+1}\cdot\ln\left [1+\frac{1}{\ln (n+1)}\right]}-1}{\ln\frac{\ln (n+1)}{\ln n}}=$](http://latex.artofproblemsolving.com/e/4/b/e4be34c28f8d04f977c34a6d4eddb0444a41d782.png)
![$\lim_{n\to\infty}\frac{\frac{1}{n+1}\cdot\ln\left [1+\frac{1}{\ln (n+1)}\right]}{\ln\frac{\ln (n+1)}{\ln n}}=$](http://latex.artofproblemsolving.com/3/f/b/3fb2823c7d97a7ce5e226b885e810c28998f1241.png)
![$\lim_{n\to\infty}\frac{\frac{1}{n+1}\cdot\frac{1}{\ln (n+1)}}{\ln\left [1+\frac{\ln (n+1)-\ln n}{\ln n}\right]}
=$](http://latex.artofproblemsolving.com/d/7/3/d732af165ba65d945f00ce68201f74b66cc4f730.png)
.
Ascertain 
, where ![$I=\left(0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/a/8/a/a8abc00f9ba0fb210e7d7ee466112d4d2d4d4871.png)
and the correspondence is 
.Proof. I will find
studyng monotony of ![$f(x)=\frac{\sin x-x\cos x}{x-\sin x},x\in \left(0,\frac{\pi}{2}\right]$](http://latex.artofproblemsolving.com/2/f/b/2fbfeb2a4946a032b40bb8d1e457e864f0db52f5.png)
. Thus 
![$\left[x-(1-\cos x)\cdot \frac{1+\sqrt{5+4\cos x}}{2\sin x}\right] \equiv\ g(x)$](http://latex.artofproblemsolving.com/2/6/9/2699852ba5c6dd63a42f0c7645937338eb3e48c5.png)
. So 
and 
. Thus, 
and 
![$\boxed {\ x\in \left(\ 0\ ,\ \frac {\pi}{2}\ \right]\ \implies\ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2\ \implies\ \mathrm{Im}(f)=\left[\frac {2}{\pi -2},2\right)\ }$](http://latex.artofproblemsolving.com/5/9/d/59d6b119ef5c7d4f16bdb88d0b874c752be7ed7b.png)
.Geometrical interpretation. Let
such that 
and 
for which 
and 
. For 
, where 
and 
, let 
so that
and 
. I"ll seek position of the remotest (to left) 
(if it exists) such that for any , have the length 
. Thus, 
, i.e. 
. Therefore, the length 
for any ![$x\in\left(0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/c/b/9/cb9b8254c4dd909eac5d5dc6affd8c6ab5c35f69.png)
. With other words, the left remotest 
for which![$\left(\forall\right) x\in\left(0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/a/8/a/a8af00d00d87ceb0b67dc9dafff2625d4115ff78.png)
we have the length of 
is 
, i.e. 
because 
is decresing and 
and 
.Applications. Prove that the following inequalities :
![$x\in \left(0,\frac{\pi}{2}\right]\Longrightarrow\frac{2}{\pi}\le\boxed {\left(1-\frac{2}{\pi}\right)\cos x+\frac{2}{\pi}\le \frac{\sin x}{x}<\frac{2+\cos x}{3}}\le 1\ ;$](http://latex.artofproblemsolving.com/7/f/1/7f106e00a7ec3db546f52d6bce000625d5aa373f.png)
.Remark. I used the equivalence relation ("same signature") over
Compute 
, where 
.Proof. The given limit can be written as
, where 
.
. I used the remarkable limits 
, 
and 
.
.I used the remarkable limits
and 
. In conclusion, 
.
Proposed problem. Find 
so that 
.Proof. I"ll use only the remarkable limit
, where 
. Observe that 
. In conclusion, our limit
and in this case 
.
PP. 
![$L_1=\lim_{x\to 0}\, \frac {\text{e}^{\frac {\ln (1+ax)}x}-\text{e}^a}x=\lim_{x\to 0}\, \left[\text{e}^a\cdot\frac {\text{e}^{\frac {\ln (1+ax)}x-a}-1}{\frac {\ln (1+ax)}x-a}\cdot\frac {\ln\left(1+ax\right)-ax}{x^2}\right]\stackrel{\text{l'H}}{=}$](http://latex.artofproblemsolving.com/7/8/4/784ad9c2d1a92a3cf4d26a97c4c2ccbb78492329.png)
![$L_2=\lim_{x\to 0}\, \frac {\text{e}^{\frac {a\ln (1+x)}x}-\text{e}^a}x=\lim_{x\to 0}\, \left[\text{e}^a\cdot\frac {\text{e}^{\frac {a\ln (1+x)}x-a}-1}{\frac {a\ln (1+x)}x-a}\cdot\frac {a\ln\left(1+x\right)-ax}{x^2}\right]\stackrel{\text{l'H}}{=}$](http://latex.artofproblemsolving.com/e/d/a/eda3d441c2d40e5e7937da16783d45b9f1114921.png)
.
PP. Ascertain 
, where 
.Proof.
.I used the remarkable limits
and 
.
PP. Ascertain 
, where 
.Proof.
, where 
.
PP. Solve recurrence system 
, , where 
.Proof. Define for any
the matrix 
, where 
and 
where 
.Observe that
for any and 
, where 
and 
.Generally, for any
we have 
, where and 
. Therefore,the associated equation
has the roots 
. Thus, for any we have
and 
and we can ascertain easily these four constans 
,where
from the first two terms of each from these two sequencies.
PP. Using eventually the well-known limit 
prove that
, where ![$a_n=\frac {\sqrt[n]{n!}}{n}$](http://latex.artofproblemsolving.com/0/6/8/06820a0dbd99d51537b0e036ac49ce0431c59ba8.png)
. Deduct that ![$\lim_{n\to\infty}\frac {n}{\ln n}\cdot\left(e-\frac {n}{\sqrt [n]{n!}}\right)=\frac e2$](http://latex.artofproblemsolving.com/c/e/f/cef8973b9bfa541238d38c06fbd3d47ab2bd9735.png)
.Proof.
, where 
,
and 
. Therefore, 
. Thus, ![$\lim_{n\to\infty}\frac {n}{\ln n}\cdot\left(e-\frac {n}{\sqrt[n]{n!}}\right)=$](http://latex.artofproblemsolving.com/3/e/2/3e2fa090577cb1506572edf4c32f7ccba08a9627.png)
![$\lim_{n\to\infty}\frac {n}{\ln n}\cdot\left(\frac {\sqrt[n]{n!}}{n}-\frac 1e\right)\cdot\frac {ne}{e\sqrt[n]{n!}-n}\cdot\frac {e\sqrt[n]{n!}-n}{\sqrt[n]{n!}}=$](http://latex.artofproblemsolving.com/3/e/e/3eefba40682899fb393acba9880b63bd09879d7b.png)
![$\frac {1}{2e}\cdot\lim_{n\to\infty}\frac {ne}{\sqrt[n]{n!}}=\frac e2$](http://latex.artofproblemsolving.com/a/c/a/aca4ba31c8fc08ebe1a54c37c228b0a4198b61d8.png)
.
PP. Let 
be the root of 
, where 
, 
. Prove that 
.Proof. Prove easilky that
for any and 
, i.e. ![$x_n > \sqrt[n]{p+1}$](http://latex.artofproblemsolving.com/c/f/5/cf5af3d30ad4d0cb5e5ebfec33297459a47e4dc9.png)
. We can set similarly 
,i.e.
![$x_n < \sqrt[n-p]{p+1}$](http://latex.artofproblemsolving.com/9/5/7/9573bdca2f3eba7629896b493a62525f35a305a6.png)
. Thus, ![$n\left(\sqrt[n]{p+1} - 1\right) < n\left(x_n - 1\right) < n\left(\sqrt[n-p]{p+1} - 1\right)$](http://latex.artofproblemsolving.com/1/2/5/125de158179c9148728390edbb94a5d3f39c6656.png)
and from the squeeze theorem, the limit is 
.
PP. Evaluate ![$\lim_{n\to \infty} \frac{n^2}{\ln n}\cdot \left(\sqrt[n]{n}-\sqrt[n+1]{n+1}\right)$](http://latex.artofproblemsolving.com/9/3/a/93acf477a8c4bbc8f99955f80f7666d0092b757f.png)
.Proof.
![$\lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot \left(\sqrt[n]{n}-\sqrt[n+1]{n+1}\right)=$](http://latex.artofproblemsolving.com/5/5/4/55408e413e08851f020bc898a2e4d3588a6d93ed.png)
![$\lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot \sqrt [n+1]{n+1}\cdot\left(\frac {\sqrt[n]n}{\sqrt[n+1]{n+1}}-1\right)=$](http://latex.artofproblemsolving.com/f/5/3/f53ee98a924d6b5a676ef2347d55701a8c4e092f.png)
![$\lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot\frac {\frac {\sqrt[n]n}{\sqrt[n+1]{n+1}}-1}{\ln\frac {\sqrt[n]n}{\sqrt[n+1]{n+1}}}\cdot \left[\frac 1n\cdot\ln n-\frac {1}{n+1}\cdot\ln (n+1)\right]=$](http://latex.artofproblemsolving.com/c/2/e/c2e4800b78e5d9feea5ab601dc205fe881fcd127.png)
.In conclusion,
![$\boxed{\ \lim_{n\to\infty}\ \frac{n^2}{\ln n}\cdot \left(\sqrt[n]{n}-\sqrt[n+1]{n+1}\right)=1\ }$](http://latex.artofproblemsolving.com/6/9/b/69b14821da6de474bd1686cc2fd5e415b6fc223e.png)
. I used the remarkable limits ![$\lim_{n\to\infty}\ \sqrt [n]n=1$](http://latex.artofproblemsolving.com/3/6/d/36de8bf895ee5ffcabd891430c63b04c17347b0b.png)
, 
and 
.
Calculate ![$\lim_{x\to\infty}\ x\cdot\left[\left(\frac {a^{\frac 1x}+b^{\frac 1x}}{2}\right)^x-\sqrt{ab}\right]$](http://latex.artofproblemsolving.com/0/a/8/0a8a932ceef259e3cfc8297c29917c79f70cd48a.png)
, where 
.Proof. Observe that
, where 
.![$\lim_{x\to\infty}\ x\cdot\left[\left(\frac {a^{\frac 1x}+b^{\frac 1x}}{2}\right)^x-\sqrt{ab}\right]\stackrel{(\infty\cdot 0)}{\ =\ }$](http://latex.artofproblemsolving.com/4/b/0/4b0552baebb528ed231bb54bfc812857af079822.png)
![$\lim_{x\to 0}\ \frac 1x\cdot\left[\left(\frac {a^x+b^x}{2}\right)^{\frac 1x}-\sqrt{ab}\right]\stackrel{(\frac 00)}{\ =\ }$](http://latex.artofproblemsolving.com/9/6/1/961cfbdd2aab59b6c7e8975ca2cfbad4cc4cc90e.png)
![$\sqrt{ab}\cdot \lim_{x\to 0}\ \frac 1x\cdot\left[\frac 1x\cdot\ln\frac {a^x+b^x}{2}-\frac 12\cdot\ln (ab)\right]=$](http://latex.artofproblemsolving.com/a/0/4/a0484d9899cb366ac17e01f9e1d6414366155597.png)
![$\frac {\sqrt{ab}}{2}\cdot\lim_{x\to 0}\ \frac {1}{x^2}\cdot\left[2\cdot\ln\frac {a^x+b^x}{2}-x\ln (ab)\right]\stackrel{(l'H)}{=}$](http://latex.artofproblemsolving.com/0/b/1/0b1c351ffeaf7d7170be6a4dfb0dc5c4a2815068.png)
![$\frac {\sqrt{ab}}{2}\cdot\lim_{x\to 0}\ \frac {1}{2x}\cdot\left[\frac {2}{a^x+b^x}\cdot\left(a^x\cdot\ln a+b^x\cdot\ln b\right)-\ln (ab)\right]=$](http://latex.artofproblemsolving.com/c/5/4/c5407a66011a128c49a49f99b5b01632e65c6013.png)
![$\frac {\sqrt{ab}}{8}\cdot\lim_{x\to 0}\ \frac 1x\cdot\left[2\left(a^x\ln a+b^x\ln b\right)-\left(a^x+b^x\right)(\ln a+\ln b)\right]=$](http://latex.artofproblemsolving.com/3/9/b/39b9a5449adb875199e1d5543772b483be3088e7.png)
![$\boxed{\ \lim_{x\to\infty}\ x\cdot\left[\left(\frac {a^{\frac 1x}+b^{\frac 1x}}{2}\right)^x-\sqrt{ab}\right]=\frac {\sqrt{ab}}{8}\cdot\ln^2\frac ab\ }$](http://latex.artofproblemsolving.com/5/9/9/599033a277a919cf98dc4c14fa408b102a96b0c8.png)
.
Without using L'Hospital's Rules, find 
.Proof. Denote
. Prove easily that 
. If 
(it existsand it is finite), then for
in the relation 
obtain that 
, i.e. 
. Therefore, 
. In conclusion, 
.II. Siruri recurente.
Articol din G.M.B. - o clasa de siruri recurente (<== click).
Examples. Ascertain the limits of the following strings 
, where 
:1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
!2.
13.
14.
15.
16.
17.
18.
19.
![$x_1>1,\ x_{n+1}=\sqrt [3] {3x_n-2},\ y_n=n(x_n-1)\ .$](http://latex.artofproblemsolving.com/3/4/4/344a8ccb25f7ef8fe2046f2120a11c3c3e26974e.png)
20.
See here and here
This post has been edited 197 times. Last edited by Virgil Nicula, Aug 16, 2017, 8:16 PM
August 2018
January 2018