Find $\lim_{x\to 0^+} (\sin x)^{(1/\ln(x))}$,without L' Hospital (homework)

Question

I got this home work assignment to find the limit without L'Hospital's rule. I used $e$ like that: $$\sin x^{1/\ln(x)} = \exp\left(\frac{\ln \sin (x)}{\ln(x)}\right)$$ and now I want to find this limit: $\lim_{x\to 0^+}\frac{\ln\sin(x)}{\ln(x)}$, but I couldn't get it through.

Answer 1

Note that for $x\to 0^+$

$$\Large{\sin x^{\frac1{\log x}}=e^{\frac{\log \sin x}{\log x}}=e^{\frac{\log \frac{\sin x}{x} +\log x}{\log x}}\to e^1=e}$$

indeed

$${\frac{\log \frac{\sin x}{x} +\log x}{\log x}}={\frac{\frac{\log \frac{\sin x}{x}}{\log x} +1}{1}}\to 1$$

Answer 2

$$\sin(x)^{1/\log(x)}=\exp\left(\log(\sin(x))*\frac{1}{\log(x)}\right)$$ $$=\exp\left(\frac{\log(\sin(x))}{\log(x)}\right)$$ Now we can use that $\sin(x)=x+O(x^2)$: $$\exp\left(\frac{\log(x+O(x^2))}{\log(x)}\right)\to\exp\left(\frac{1}{1}\right)=e$$

Answer 3

The simplest is with equivalents:

$\sin x\sim_0 x$, hence $$\ln(\sin x)\sim_0\ln x,\enspace\text{so }\enspace \sim_0\frac{\ln x}{\ln x}=1,$$ whence $\;\displaystyle\lim_{x\to 0}\,\mathrm e^{\tfrac{\ln(\sin x)}{\ln x}}=\mathrm e^1=\mathrm e$.