without L'Hospital's rule find the $\lim_{x\to 0+} (\csc x)^{\sin^2 x}$

Question

$\lim_{x\to 0+}(\csc x)^{\sin^{2}x}$

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,.

Answer 1

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

Answer 2

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 $$ \lim_{x\to0^+}(\csc x)^{\sin^2x}=1 $$

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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.