$\lim_{x\to 0+}=(1/x)^{\sin x}$?

Question

$\lim_{x\to 0+}=(1/x)^{\sin x}$

I think I should rewrite it into a from $e^{\ln}$ , but I can't continue the calculation after this step.

Answer 1

HINT:

$$\lim_{x\to 0}\left(\frac{1}{x}\right)^{\sin(x)}=$$ $$\lim_{x\to 0}\exp\left(\ln\left(\frac{1}{x}\right)\sin(x)\right)=$$ $$\exp\left(\lim_{x\to 0}\ln\left(\frac{1}{x}\right)\sin(x)\right)=$$ $$\exp\left(\lim_{x\to 0}-\ln(x)\sin(x)\right)=$$ $$\exp\left(\lim_{x\to 0}\frac{-\ln(x)}{\frac{1}{\sin(x)}}\right)=$$ $$\exp\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(-\ln(x)\right)}{\frac{\text{d}}{\text{d}x}\frac{1}{\sin(x)}}\right)=$$ $$\exp\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(-\ln(x)\right)}{\frac{\text{d}}{\text{d}x}\csc(x)}\right)=$$ $$\exp\left(\lim_{x\to 0}\frac{\sin^2(x)}{x\cos(x)}\right)=$$ $$\exp\left(\lim_{x\to 0}\frac{1}{\cos(x)}\cdot\lim_{x\to 0}\frac{\sin^2(x)}{x}\right)=$$ $$\exp\left(\frac{1}{\cos(0)}\cdot\lim_{x\to 0}\frac{\sin^2(x)}{x}\right)=$$ $$\exp\left(\lim_{x\to 0}\frac{\sin^2(x)}{x}\right)$$

Answer 2

Hint

$$\left(\frac{1}{x}\right)^{\sin(x)}=e^{-\sin(x)\ln(x)}=e^{-\frac{\sin x}{x}\cdot x\ln(x)}.$$

Answer 3

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

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