Calculate $\lim_{x\to 0}\frac{1}{x^{\sin(x)}}$

Question

Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$

I'm pretty much clueless here, only that there is L'hospital obviously here.

Would appreciate any help.

Answer 1

It usually helps to consider the log of the expression in such limits:

$$\log{\left [ x^{-\sin{x}} \right ]} = -\sin{x} \log{x} $$

$$\lim_{x \rightarrow 0} \sin{x} \log{x} = \lim_{x \rightarrow 0} x \log{x} = 0 $$

The limit in question, therefore, is 1.

EDIT

I can be a little more clear on using L'Hopital on the above limit:

$$\lim_{x \rightarrow 0} x \log{x} = \lim_{x \rightarrow 0} \frac{\log{x}}{1/x} $$

Now use L'Hopital:

$$ \lim_{x \rightarrow 0} \frac{\log{x}}{1/x} = \lim_{x \rightarrow 0} \frac{1/x}{-1/x^2} = \lim_{x \rightarrow 0} (-x) $$

Answer 2

$$ x^{-\sin x} = \left( e^{\ln(x) } \right) ^{-\sin x} = e^{-\sin x \ln (x)}$$

Since this is of indeterminate from, we can apply L'hopital rule here. $$ \large \lim_{x \to 0} x^{-\sin x} = e^{\lim_{x \to 0} \left( \frac{-\sin x}{\frac{1}{\ln x}} \right)}$$

Answer 3

Things are simple if we use the elementary limits $\lim_{x\to 0}x^x=1$ and$\lim_{x\to 0}\frac{\sin x}{x}=1$ $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}=\lim_{x\to 0} x^{\displaystyle x\frac{\sin(x)}{x} (-1)}=1$$

Q.E.D.