[/quote]
,
,
,
cut three common chords concurrent at 
.
,
which are internally tangent to all 
.
be a parallelogram of center 
.
,
and 
such that 
.
be given . On the extension of the segment 
beyond 
,
is chosen arbitrarily. Let the line through 
intersect the line 
at the point 
.
be the intersection of the lines 
and 
.
is tangent to the circumcircle of the triangle 
.
be any point. Let 
be projection point from 
.
cuts circumcircle 
at 
.
be antipode of 
are cyclic.
,
be real numbers, such that 
.
.
be a distribution function. Prove that for any 
,![\[
G(x) = \frac{1}{2h} \int_{x-h}^{x+h} F(t) \, dt
\]](http://latex.artofproblemsolving.com/5/5/d/55d81a7d36b0a052ddeb3e189645e43eb41f61a5.png)
.
such that 
.
passing trough 
at 
.
.
of the equation 
which follows 
(b) Solve for real 
Compute -![\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}](http://latex.artofproblemsolving.com/8/6/f/86fb48db976012106bd30f74cb6ada8c6e6d7800.png)
For 
define 
Determine the open interval(s) (if any) where 
is decreasing and all the open interval(s) (if any) where 
Determine all the local minima of 
where 
Y by
Solve:
\lim_{ x \to \frac{\pi}{2}^+ } tan(x)
NB:The calculus textbook I'm reading gives the answer
as as ( -\infty ) and not '0.027'.
( The textbook doesn't provide any algebraic justification
for this answer, it just plots the graphs.
But i'll like a Clear algebraic explanation
)
\lim_{ x \to \frac{\pi}{2}^+ } tan(x)
NB:The calculus textbook I'm reading gives the answer
as as ( -\infty ) and not '0.027'.
( The textbook doesn't provide any algebraic justification
for this answer, it just plots the graphs.
But i'll like a Clear algebraic explanation
)
This post has been edited 1 time. Last edited by deepthinka, Mar 19, 2025, 9:12 PM
![\[\tan x=\frac{\sin x}{\cos x},\,\lim_{x\to\frac\pi2^+}\sin x=1,\,\lim_{x\to\frac\pi2^+}\cos x=0\]](http://latex.artofproblemsolving.com/8/2/9/8296bdfbf005f740344e6415b487ca54d4660b22.png)
thus![\[\tan x\to\frac10\text{ as }x\to\frac\pi2^+\]](http://latex.artofproblemsolving.com/7/c/a/7caac45d2e7c346bc90f0a7f8ee6442407434f7f.png)
so we analyze the behavior of 
and 
instead. 
and 
as 
so we have ![\[\lim_{x\to\frac\pi2^+}\tan x=\lim_{x\to\frac\pi2^+}\frac{\sin x}{\cos x}=\lim_{u\to0^-}\frac1u=-\infty\]](http://latex.artofproblemsolving.com/9/0/6/9069435b10ba2d4c2f210fcfd56d5b34b188c38e.png)
(not rigorous, but 
exhibits similar behavior as 
similar to how 
does as 
)
then ![\[\tan x=\tan(\pi/2-u)=\cot u.\]](http://latex.artofproblemsolving.com/a/f/f/affaebd0289b7e6bf28bb61c79d602b126cd0713.png)
thus ![\[\lim_{x\to\frac\pi2^+}\tan x=\lim_{u\to0^-}\cot u=\lim_{u\to0^-}\frac{\cos u}{\sin u}.\]](http://latex.artofproblemsolving.com/b/e/1/be17f5548d0c085c18c1915467dccd06bd3ccb77.png)
now, use the small angle formulae: ![\[\lim_{u\to0^-}\frac{\cos u}{\sin u}=\lim_{u\to0^-}\frac{1-\frac12u^2}u=\lim_{u\to0^-}\frac1u-\lim_{u\to0^-}\frac12u=\lim_{u\to0^-}\frac1u=-\infty.\]](http://latex.artofproblemsolving.com/2/b/b/2bb1ad479f1131c7bcdb2c3831a8aea79e496245.png)
so, ![\[\lim_{x\to\frac\pi2^+}\tan x=-\infty.\]](http://latex.artofproblemsolving.com/1/9/9/199ce2f85fe5d821be50adec4cca71dd5dd95794.png)
(this method wasn't completely rigorous, but it was intuitively understandable.) or, go with the book and look at the graph! as 
it zooms downward, so its limit is understandably 
