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  • Let <img alt="\mathrm{f(x)=\lim _{n \rightarrow \infty}(\sin x)^{2 n}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%28%5

Let \mathrm{f(x)=\lim _{n \rightarrow \infty}(\sin x)^{2 n}}; then f is

Option: 1

continuous at \mathrm{x=\frac{\pi}{2}}


Option: 2

continuous at \mathrm{x=\frac{\pi}{2}}


Option: 3

discontinuous at \mathrm{x=-\frac{\pi}{3}}


Option: 4

discontinuous at infinite number of points.


Answers (1)

best_answer

\mathrm{\text { Since } \lim _{n \rightarrow \infty} x^{2 n}= \begin{cases}0, & \text { if }|x|<1 \\ 1, & \text { if }|x|=1\end{cases}}

\mathrm{\therefore f(x)=\lim _{x \rightarrow-}(\sin x)^{2 n}= \begin{cases}0 & \text { if }|\sin x|<1 \\ 1 & \text { if }|\sin x|=1\end{cases}}

Thus, f(x) is continuous at all x, except for those values of x for

which \mathrm{|\sin x|=1 ~i.e. ~x=(2 k+1) \frac{\pi}{2}, where ~k \in Z}
For these points \mathrm{\lim _{z \rightarrow(2 k+1) \frac{\pi}{2}} f(x)=0 \neq 1=f\left((2 k+1) \frac{\pi}{2}\right)}

Hence, f(x) is discontinuous at these points.

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manish

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