1. ## Limit 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)

2. Please use proper grouping symbols so we know exactly what you mean.

And never spell 'pi' as 'pie'. This is a math forum, not a bakery.

Anything to the 0 power is 1.

$\displaystyle \lim_{x\rightarrow{0^{+}}}[cos(\frac{\pi}{2}-x)]^{x}=1$.

3. Originally Posted by galactus
$\displaystyle \lim_{x\rightarrow{0^{+}}}[cos(\frac{\pi}{2}-x)]^{x}=1$.
Are you sure?

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:

$\displaystyle \lim_{u\rightarrow{1}}\frac{1-u}{u^{4}+2u^{2}-3}$

$\displaystyle \lim_{u\rightarrow{1}}\frac{-1}{u^{3}+u^{2}+3u+3}=\boxed{\frac{-1}{8}}$

5. Originally Posted by Krizalid
Are you sure?

I checked it with my TI-92 and got this.

6. 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: .$\displaystyle \lim_{x\to0^+} y \;=\;e^0 \;=\;1$