I have no clue what to do here....please help
lim as x goes to 0+ of [cos(pi/2-x)]^x
and
lim as x goes to 2+ of (1/x^2-4 - sqr(x-1)/(x^2-4)
Rewrite $\displaystyle \frac{1}{x^{2}-4}-\frac{\sqrt{x-1}}{x^{2}-4}$
as $\displaystyle \frac{1-\sqrt{x-1}}{x^{2}-4}$
Now, make the sub: $\displaystyle u=\sqrt{x-1}$
You get:
$\displaystyle \lim_{u\rightarrow{1}}\frac{1-u}{u^{4}+2u^{2}-3}$
$\displaystyle \lim_{u\rightarrow{1}}\frac{-1}{u^{3}+u^{2}+3u+3}=\boxed{\frac{-1}{8}}$
Hello, poopforbrains!
I found a solution.
. . I hope someone has a better way!
$\displaystyle \lim_{x\to0^+}\,\left[\cos\left(\frac{\pi}{2} - x\right)\right]^x$
Note that: .$\displaystyle \cos\left(\frac{\pi}{2} - x\right) \;=\;\sin(x)$
So we have: .$\displaystyle y \;=\;\left[\sin(x)\right]^x$
Take logs: .$\displaystyle \ln(y) \;=\;\ln\left[\sin(x)\right]^x \;=\;x\cdot\ln[\sin(x)] \;=\;\frac{\ln[\sin(x)]}{\frac{1}{x}}$ . . . which approaches $\displaystyle \frac{-\infty}{\infty}$
. . Apply L'Hopital: .$\displaystyle \frac{\dfrac{\cos x}{\sin x}}{-\dfrac{1}{x^2}} \;=\;- \frac{x^2}{\tan x}$ . . . which approaches $\displaystyle \frac{0}{0}$
. . Apply L'Hopital: .$\displaystyle -\frac{2x}{\sec^2\!x}$ . . . which approaches $\displaystyle -\frac{0}{1} \:=\:0$
Hence: .$\displaystyle \lim_{x\to0^+} \ln(y)\;=\;0$
Therefore: .$\displaystyle \lim_{x\to0^+} y \;=\;e^0 \;=\;1$