How to find this limit with Landau's symbols?

Question

$$\lim_{x\to0^+} \left(\frac{x^{\sin(x)}-1}{x}\right)$$

I tried putting $1+o(1)$ in place of $x^{\sin(x)}$ which gives me $\frac{o(1)}{x}$ but i don't know where to go from here. And I also don't know how to find the limit $x^{\sin(x)}$. I broke it down to $x^{x+o(x)}$ which points towards the limit being $1$.

Answer 1

Since $$\sin{x}\ln{x}=\frac{\sin{x}}{x}\cdot\ln{x^x}\rightarrow0,$$ we obtain: $$\frac{x^{\sin{x}}-1}{x}=\frac{e^{\sin{x}\ln{x}}-1}{\sin{x}\ln{x}}\cdot\frac{\sin{x}}{x}\cdot\ln{x}\rightarrow-\infty.$$

Answer 2

Considering $$A=\frac{x^{\sin(x)}-1}{x}$$ start considering $$B=x^{\sin(x)}\implies \log(B)=\sin(x)\log(x)$$ Using Taylor expansion of $\sin(x)$,we then have $$\log(B)=x \log (x)-\frac{1}{6} x^3 \log (x)+O\left(x^5\right)$$ Now, continuing with Taylor $$B=e^{\log(B)}=1+x \log (x)+\frac{1}{2} x^2 \log ^2(x)+\frac{1}{6} x^3 \log (x) \left(\log ^2(x)-1\right)+O\left(x^4\right)$$ Then $$A=\frac{B-1}x=\log (x)+\frac{1}{2} x \log ^2(x)+\frac{1}{6} x^2 \log (x) \left(\log ^2(x)-1\right)+O\left(x^3\right)$$