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  • <img alt="\lim _{x \rightarrow 0^{+}}(\operatorname{cosec}(x))^{\sin ^2(x)}" src="/latex-image/?%5Clim%20_%7Bx%20%5Crightarrow%200%5E%7B&plus;%7D%7D%28%5Coperatorname%7Bcosec%7D%28x%29%29%5E%7B

\lim _{x \rightarrow 0^{+}}(\operatorname{cosec}(x))^{\sin ^2(x)}

Option: 1

1


Option: 2

3


Option: 3

0


Option: 4

2


Answers (1)

best_answer

Let's see where you went wrong. We know that

                                                    \mathrm{ \begin{aligned} & \lim _{x \rightarrow 0^{+}} \sin (x)=0 \\\\ & \Longrightarrow \lim _{x \rightarrow 0^{+}} \operatorname{cosec}(x)=+\infty \\\\ & (\because \sin (x)>0 \forall x \in(0, \pi)) \\\\ & \Longrightarrow \lim _{x \rightarrow 0^{+}} \ln (\operatorname{cosec}(x))=+\infty \\ & \end{aligned} }
And you claimed that to be 0 .

Now, coming to the question, let the required limit be y.

                                           \mathrm{ \begin{gathered} \Longrightarrow y=\lim _{x \rightarrow 0^{+}}(\operatorname{cosec}(x))^{\sin ^2(x)} \\\\ \Longrightarrow \ln (y)=\lim _{x \rightarrow 0^{+}}\left(\sin ^2(x)\right) \ln (\operatorname{cosec}(x)) \end{gathered} }

As \mathrm{x \rightarrow 0^{+}, \operatorname{cosec}(x) \rightarrow \infty}. Hence,

                                   \mathrm{ \begin{gathered} \Longrightarrow \ln (y)=\lim _{u \rightarrow \infty} \frac{\ln (u)}{u^2} \\\\ \Longrightarrow \ln (y)=\lim _{u \rightarrow \infty} \frac{\ln (u)}{u} \times \lim _{u \rightarrow \infty} \frac{1}{u} \end{gathered} }

both of which limits are clearly 0 .

                                          \mathrm{ \begin{gathered} \Longrightarrow \ln (y)=0 \times 0=0 \\\\ \Longrightarrow y=1 \end{gathered} }

Posted by

avinash.dongre

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