# Use L'Hopital's with this problem?

The problem is: $$\lim_{x\rightarrow 0^+} \left(\frac{1}{x}\right)^{\sin x}$$ I know the answer is $1$ because I checked with my graphing calculator, but how exactly do I get there?

I got this far: $$\ln y= \lim_{x\rightarrow0^+} \frac{ \ln\left(\frac{1}{x}\right)}{\csc x}$$

Then I used L'Hop and got to the form: $$\ln y = \lim_{x\rightarrow0^+} \frac{1}{\csc x \cot x}$$ The problem is that when I plug in the $0$ into the $x$'s, I get $1/ (0)(\infty)(\infty)$. But isn't $(0)(\infty)$ an indeterminate form...?

How can I solve this limit so that I can get $\lim_{x\rightarrow0} = 0$, raise it to $e$ (to get rid of $\ln$) and get my final answer of $1$?

Thank you.

Let $$\displaystyle y=\lim_{x\rightarrow 0^{+}}\left(\frac{1}{x}\right)^{\sin x}\;,$$ Now Let $x=0+h\;,$ Then $\displaystyle y=\lim_{h\rightarrow 0}\left(\frac{1}{h}\right)^{\sin h}$

So $$\displaystyle \ln(y) = \lim_{h\rightarrow 0}\sin (h)\cdot \ln\left(\frac{1}{h}\right) = -\lim_{h\rightarrow 0}\sin h\cdot \ln(h) = -\lim_{h\rightarrow 0}\frac{\ln(h)}{\csc (h)}\left(\frac{\infty}{\infty}\right)$$ form

Now Using $\bf{D-LHopital \; Rule}$

$$\displaystyle \ln(y) = \lim_{h\rightarrow 0}\frac{1}{h\cdot \csc h\cdot \cot h}=\lim_{h\rightarrow 0}\frac{\sin^2 h}{h\cdot \cos h} = \lim_{h\rightarrow 0}\frac{\sin h}{h}\times \lim_{h\rightarrow 0}\frac{\sin h}{\cos h} = 1\times 0$$

So We get $$\ln(y) = 0\Rightarrow y=e^{0} = 1$$

• you freaking rock!!! you're the best. Sep 3, 2015 at 1:46
• You set $x=h.$ How does that help? Also, $0^+$ became $0?$
– zhw.
Sep 3, 2015 at 1:53