# Find $\lim_{x\to\infty} x^{\sin(1/x)}$

How to find $\lim_{x\to\infty} x^{\sin(1/x)}$?

I tried

$$\lim_{x\to\infty} x^{\sin(1/x)}=\lim_{x\to\infty}e^{\sin(1/x)\ln(x)}$$

Then

$$\lim_{x\to\infty}\sin\left(\frac{1}{x}\right)\ln(x)=\lim_{x\to\infty}\frac{\sin(1/x)}{\frac{1}{\ln(x)}}=\lim_{x\to\infty}\frac{\cos(1/x)}{x^2}x\ln^2(x)=\lim_{x\to\infty}\frac{1}{x}\cos\left(\frac{1}{x}\right)\ln^2(x)$$

Which doesn't look promising.

• In the last step just take the upper and lower bound on $\cos$ and show that it converges to 0. Hence your expression converges to 1. – Alex Sep 11 '16 at 22:28

## 7 Answers

You're on the right track. We have $\lim_{x\to \infty}\cos(1/x) = \lim_{y\to 0}\cos y = 1$, and $\lim_{x\to \infty} (1/x)\ln^2 x = 0$. Indeed, for all $x \ge 1$,

$$0 < \frac{\ln^2 x}{x} = \frac{16\ln^2(\sqrt[4]{x})}{x} \le \frac{16\sqrt{x}}{x} = \frac{16}{\sqrt{x}}$$

and the squeeze theorem yields $\lim_{x\to \infty} (1/x)\ln^2 x = 0$. So $\lim_{x\to \infty} \sin(1/x)\ln x = 0$, and consequently $\lim_{x\to \infty} x^{\sin(1/x)} = 1$.

When $x$ goes to $\infty$, you have $1/x\to 0$, and $\sin(X)\sim_{x\to 0}X$.

So $$\sin(1/x)\ln(x)\sim \frac{\ln(x)}x\xrightarrow[x\to\infty]{} 0.$$

Finally, because you can compose equivalents with $\exp$ you get the limit $e^0=1$.

So

$$x^{\sin(1/x)}\xrightarrow[x\to\infty]{} 1.$$

After taking logs and writing $h = \frac{1}{x}$, this is the same as $$\lim_{h \to 0} \sin h \cdot \ln \frac{1}{h} = -\lim_{h \to 0} \sin h \cdot \ln h = -\lim_{h \to 0} \frac{\sin h}{h} \cdot \lim_{h \to 0} h \cdot \ln h\, .$$ Both limit should be familiar to you.

$$\lim_{x\to\infty} x^{(1/x)}=1$$ $$\sin(1/x)<= 1/x$$ so the limit is $1$

Hint: Our expression equals

$$(x^{1/x})^{\sin(1/x)/(1/x)}.$$

Recall $x^{1/x} \to 1.$

Use L'hopital's twice to solve the last limit. First note that $\lim_{x \to \infty} \cos(1/x)=\lim_{x \to 0}\cos(x)=1$. Then, we can easily get that

$$0 \leq \lim_{x\to\infty}\frac{1}{x}\cos\left(\frac{1}{x}\right)\ln^2(x)\leq \lim_{x \to \infty} \frac{\ln^2(x)}{x}=\lim_{x \to \infty} \frac{2\ln(x)}{x^2}=\lim_{x \to \infty}\frac{1}{x^2}=0$$

By the squeeze theorem the limit is $0$, implying the result, since $e^0=1$.

$$\lim_{x\to\infty}\sin\left(\frac{1}{x}\right)\ln(x) = \lim_{x\to\infty}\frac{\ln(x)}{\csc(\frac{1}{x})}=\lim_{x\to\infty}\frac{1/x}{(\csc \frac{1}{x} \cot \frac{1}{x})/x^2} = \lim_{x\to\infty} \frac{\sin (1/x)}{1/x} \frac{\tan (1/x)}{1/x} \frac{1}{x} = 0$$