I have this limit
$$\lim_{x\to 0}(1-x)^{\frac{\log\left(1+\frac{x^2}2\right)}{\sin^5x}} \tag 1$$
With the Hôpital's rule written with the form
$$e^{g(x)\log f(x)}$$
I will have the indeterminate form $\left(\frac00\right)$ if $x\to 0$. I have thinked to write the $(1)$ as:
$$\left[\left(1+\frac{1}{-\frac 1x}\right)\right]^{-\frac 1x}\to e$$ with $t=-1/x$. Infact if $x\to 0 \implies t\to \infty$. Considering only the exponent, I will have $$ \frac{-x\log(1+x^2/2)}{\sin^5 x}=\frac{x}{\sin x}\cdot \frac{-\log(1+x^2/2)}{\sin^4x}\to_0 \quad 1\cdot \frac00$$
and I think now that this way it is not correct. Is there another mode to solve the $(1)$? At this moment I have not another idea.
By standard limits we have that
$$\frac{\log\left(1+\frac{x^2}2\right)}{\sin^5x}\log(1-x)=\frac{\log\left(1+\frac{x^2}2\right)}{\frac{x^2}2}\frac{x^5}{\sin^5x}\frac{\log(1-x)}{x}\frac{1}{2x^2}\to -\infty $$
therefore
$$\lim_{x\to 0}(1-x)^{\frac{\log\left(1+\frac{x^2}2\right)}{\sin^5x}}=0 $$