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Solving \mathrm{ \lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2 \cos (x)}\right)}{\sin \left(\sin \left(x^2\right)\right)}}

Option: 1

-\frac{\pi}{4}


Option: 2

\mathrm{\frac{\pi}{4}}


Option: 3

\mathrm{\frac{\pi}{2}}


Option: 4

0


Answers (1)

best_answer

                                                \mathrm{\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2 \cos (x)}\right)}{\sin \left(\sin \left(x^2\right)\right)}}

\mathrm{\text { Using the identity, } \cos (x)=\sin \left(90^{\circ}-x\right)}

                  \mathrm{\Longrightarrow \lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2 \cos (x)}\right)}{\sin \left(\sin \left(x^2\right)\right)}=\lim _{x \rightarrow 0} \frac{\sin \left(\frac{\pi}{2}-\frac{\pi}{2 \cos (x)}\right)}{\sin \left(\sin \left(x^2\right)\right)}}

                                                                \mathrm{\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\left(\frac{\pi}{2}-\frac{\pi}{2 \cos (x)}\right) \cdot \frac{\sin \left(\frac{\pi}{2}-\frac{\pi}{2 \cos (x)}\right)}{\left(\frac{\pi}{2}-\frac{\pi}{2 \cos (x)}\right)}}{\sin \left(\sin \left(x^2\right)\right)} \\ & =\lim _{x \rightarrow 0} \frac{\left(\frac{\pi}{2}-\frac{\pi}{2 \cos (x)}\right)}{\sin \left(\sin \left(x^2\right)\right)} \cdot \underbrace{\lim _{x \rightarrow 0} \frac{\sin \left(\frac{\pi}{2}-\frac{\pi}{2 \cos (x)}\right)}{\left(\frac{\pi}{2}-\frac{x}{2 \cos (x)}\right)}}_1 \end{aligned}}

                                                                   \mathrm{\begin{aligned} & =\frac{\lim _{x \rightarrow 0} \frac{\pi}{2}\left(\frac{\cos (x)-1}{\cos (x)}\right)}{\underbrace{\lim _{x \rightarrow 0} \frac{\sin \left(\sin \left(x^2\right)\right)}{\sin \left(x^2\right)}}_1 \cdot \sin \left(x^2\right)} \\ & =\frac{\lim _{x \rightarrow 0} \frac{\pi}{2}\left(\frac{\cos (x)-1}{\cos (x)}\right)}{\underbrace{\lim _{x \rightarrow 0} \frac{\sin \left(x^2\right)}{x^2}}_1 \cdot x^2} \\ & =\lim _{x \rightarrow 0} \frac{\pi}{2 x^2}\left(\frac{\cos (x)-1}{\cos (x)}\right) \\ & \end{aligned}}

                                                                       \mathrm{\begin{aligned} & \lim _{x \rightarrow 0} \frac{\pi}{2 x^2} \cdot \frac{\cos (x)-1}{\cos (x)} \\ = & \lim _{x \rightarrow 0} \frac{-(1-\cos (x))}{x^2} \cdot \frac{\pi}{2 \cos (x)} \end{aligned}}

\mathrm{\text { The limit } \lim _{x \rightarrow 0} \frac{1-\cos (x)}{x^2}=\frac{1}{2}}

                                    \mathrm{\begin{aligned} \Longrightarrow & \lim _{x \rightarrow 0} \frac{-(1-\cos (x))}{x^2} \cdot \frac{\pi}{2 \cos (x)}=\lim _{x \rightarrow 0} \frac{-1}{2} \cdot \frac{\pi}{2 \cos (x)} \\ & =\frac{-\pi}{4 \cos (0)}=\frac{-\pi}{4} \end{aligned}}

\mathrm{\text { So the limit is }-\pi / 4 \text {. }}

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Anam Khan

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