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

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.

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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:

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

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

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

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.

Here is an approach that uses neither L’Hospital’s Rule nor logarithms. Rather, it relies only on standard inequalities from geometry along with the Squeze Theorem.

To that end, we proceed by recalling that for $x\ge 0$ the sine function satisfies the inequalities

$$x\cos x\le \sin x\le x \tag 1$$

From $(1)$ it is easy to show that for $0\le x\le 1$ we have

$$x\sqrt{1-x^2} \le \sin x\le x \tag 2$$

Then, using $(2)$, we have for $0\le x\le 1$

$$\left(\frac1x\right)^{x\sqrt{1-x^2}}\le \left(\frac1x\right)^{\sin x}\le \left(\frac1x\right)^{x} \tag 3$$

Using $\lim_{x\to 0^+}x^x=1$ in $(3)$ along with the Squeeze Theorem reveals

$$\lim_{x\to 0^+}\left(\frac1x\right)^{\sin x}=1$$