without L'Hospital's rule find the $\lim_{x\to 0+} (\csc x)^{\sin^2 x}$ $\lim_{x\to 0+}(\csc x)^{\sin^{2}x}$
Without using L'Hospital's rule, I've managed to get do:
Let $y=(\csc x)^{\sin^{2}x}$ now by taking $\ln$ of both sides I got
$\lim_{x\to 0+}y$=$\lim_{x\to 0+}\sin^{2}x\ln(\csc x)$. Now by using the limit rules of a product if $\lim_{x\to 0+}\sin^{2}x=T$ and the $\lim_{x\to 0+}\ln(\csc x)=H$ then   $\lim_{x\to 0+}(\csc x)^{\sin^{2}x}=TH$
$\lim_{x\to 0+}\sin^{2}x=0$ by algebra of limits, so the overall limit should either be indeterminate form or $0$ since $0.H=0$ if H is a non-indeterminate form
Knowing this I used the epsilon-delta definition to prove $\lim_{x\to 0+}\ln(\csc x)=0$ 
Let $\epsilon$>$0$ such that
$|\ln(\csc x)-0|<\epsilon$. I can ensure that $\ln(\csc x)<\epsilon$ by requiring $x<\sin^{-1}$ $\frac{1}{e^{\epsilon}}$. Thus:
I let $\delta=$ $\sin^{-1}$ $\frac{1}{e^{\epsilon}}$. And I think the conditions of the definition were satisfied. There the $\lim_{x\to 0+}\ln(\csc x)=0$ 
Thus $\lim_{x\to 0+}(\csc x)^{\sin^{2}x}=0$
My question is where have I gone wrong because the answer is 1, but I cannot find my own fault.
Thank you in advance,.
 A: No need to use $\varepsilon$-$\delta$. See also at the end for other comments.
Note first that $\ln\csc x=-\ln\sin x$ so you have to compute
$$
\lim_{x\to0}-\sin^2x \ln\sin x
$$
Now use the substitution $x=\arcsin t$, so the limit becomes
$$
\lim_{t\to 0}-t^2\ln t
$$
which is known to be $0$. Since
$$
\lim_{x\to0^+}\ln\bigl((\csc x)^{\sin^2x}\bigr)=0
$$
you can conclude that
$$
\lim_{x\to0^+}(\csc x)^{\sin^2x}=1
$$

Note that
$$
\lim_{x\to0}\ln\csc x=\infty
$$
and certainly not $0$ as you claim. So, even if you found the correct limit $0$, your method is flawed.
A: Let's see where you went wrong. We know that
$$\lim_{x\to 0^+}sin(x)=0$$
$$\Longrightarrow \lim_{x\to 0^+}cosec(x)=+\infty$$
$$(\because sin(x)>0 \forall x\in(0,π))$$
$$\Longrightarrow \lim_{x\to 0^+}ln(cosec(x))=+\infty$$
And you claimed that to be $0$.
Now, coming to the question, let the required limit be $y$.
$$\Longrightarrow y=\lim_{x\to 0^+}(cosec(x))^{sin^2(x)}$$
$$\Longrightarrow ln(y)=\lim_{x\to 0^+}(sin^2(x))ln(cosec(x))$$
As $x\to 0^+,cosec(x)\to \infty$. Hence,
$$\Longrightarrow ln(y)=\lim_{u\to \infty}\frac{ln(u)}{u^2}$$
$$\Longrightarrow ln(y)=\lim_{u\to\infty}\frac{ln(u)}{u}×\lim_{u\to\infty}\frac{1}{u}$$
both of which limits are clearly $0$.
$$\Longrightarrow ln(y)=0×0=0$$
$$\Longrightarrow y=1$$
which is the required answer.
Hope it helps
