Limit involving $x$ raised to $\sin (1/x)$

Question

the limit as $x \to \infty$ of $x ^{\sin(1/x)}$

I see this is an infinity to zero situation, and I so I know this will be an $e^{\text{something}}$ sort of answer and that my next step would look something like: limit as $x$ goes to infinity of $\ln(x^{\sin(1/x)})$... what is my next step and how do I proceed?

Thanks much!

Answer 1

We have that

$$x ^{\sin(1/x)}=e^{\sin \left(\frac1x\right) \log x}=e^{\frac{\sin \left(\frac1x\right)}{\frac1x} \cdot \frac{\log x}{x}}\to e^{1 \cdot 0}=e^0=1$$

indeed since $t=\frac1x \to 0$

$$\frac{\sin \left(\frac1x\right)}{\frac1x}=\frac{\sin t}{t}\to 1$$

and it easy to show (for example by l'Hopital)

$$\frac{\log x}x \to 0$$

Answer 2

Alternatively, note: $\sin \frac 1x<\frac 1x, x>0$. Then: $$\left(\sin \frac 1x\right)^{\sin \frac 1x}<\left(\frac 1x\right)^{\sin \frac 1x}<x^{\sin \frac 1x}<x^{\frac 1x}, x>1$$ Taking limit at $x\to\infty$: $$1\le \lim_{x\to\infty} x^{\sin \frac 1x}\le 1.$$

Also: See this link.

Answer 3

Consider $$y=x^{\sin \left(\frac{1}{x}\right)}\implies \log(y)={\sin \left(\frac{1}{x}\right)}\log(x)$$ Now, use Taylor expansion for large $x$ (or use equivalents) $${\sin \left(\frac{1}{x}\right)}=\frac{1}{x}+O\left(\frac{1}{x^3}\right)$$ then $$\log(y) \sim \frac{\log(x)}x\to 0 \qquad \text{making}\qquad y \to 1$$