Highschool trig limit $\lim_{x\to0} (\sin(x))^{\cot^2(x) }$ $$\lim_{x\to0} (\sin(x))^{\cot^2(x) }$$
simply putting in the calculator gives not defined and seeing its graph we can clearly say its not defined but I still want to see the whole solving process. I also asked my teacher he said its not solvable.
 A: If $x<0$ then $\sin x<0.$ If $\cot^2 x$ is irrational, we are trying to take an irrational power of a negative number, which is undefined. The limit does not exist.
A: There's a standard trick here, but it gets messy if you aren't careful.
The main problem is that $\sin(x)^{\cot^2(x)}$ might not be defined for small, negative $x$. In that case, $\sin(x)$ will be negative, and you can't always raise a negative number to any power you want without doing something complicated. If we're happy using one-sided limits, then our trick will work.
If $\sin(x)^{\cot^2(x)}$ has a nonzero limit from the right at $0$, then so does
$$L(x) = \ln \sin(x)^{\cot^2(x)}.$$
This function is only defined where it makes sense. In this case that's for small, positive $x$.
Using the properties of the logarithm,
$$L(x) = \cot^2(x) \ln \sin(x) = \frac{\ln \sin(x)}{\tan^2(x)}.$$
As $x \to 0$ from the right, this is a limit of the form $-\infty / 0$, which comes out to $-\infty$ since $\tan^2(x)$ is positive near $0$. Thus,
$$\lim_{x \to 0^+} \sin(x)^{\cot^2(x)} = \lim_{x \to 0^+} e^{L(x)} = e^{-\infty} = 0.$$
So from the right, $\sin(x)^{\cot^2(x)}$ approaches $0$. But from the left it's mostly meaningless.
