Where did i go wrong in limits?

As an alternative, if you have at your disposal the standard limits ${\sin x\over x}\to1$ and $x^x\to1$ as $x\to0$, then

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

It's good to understand tricks instead of just blindly using them. I don't exactly understand the reasoning for the trick, but I believe it makes use of the taylor approximation of the logarithm centered around $1$ (i.e. when $|x-1|\ll 1$, $\ln x\approx x-1$)

Anyway, looking at all the examples in the video, I believe this trick only works when dealing with the $1^\infty$ indeterminate form and not the $\infty^0$ indeterminate form that your problem is dealing with. This makes sense as you would be applying the taylor approximation when the terms inside the natural log approach $1$. I'll add a proof if I can figure out one.

Well, I didn't prove it myself, but I found a quora post that did. I'll just rewrite the proof. Let's say we want to compute $\lim_{x\to a} f(x)^{g(x)}$ where $\lim_{x\to a} f(x)=1$ and $\lim_{x\to a} g(x)=\infty$. Then we have $$\lim_{x\to a} f(x)^{g(x)}$$ $$=\lim_{x\to a} (1+(f(x)-1))^{g(x)(f(x)-1)\cdot \frac{1}{f(x)-1}}$$ $$=\lim_{x\to a} \left((1+(f(x)-1))^{\frac{1}{f(x)-1}}\right)^{g(x)(f(x)-1)}$$ $$=\lim_{x\to a} e^{g(x)(f(x)-1)}$$ $$=e^{\lim_{x\to a} g(x)(f(x)-1)}$$