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Tell me? - Limit , continuity and differentiability - JEE Main-2

\lim_{x\rightarrow 1} \cos (\frac{\pi}{2}\sin\frac{\pi}{2} x ) equals

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As we have learned

Limits of composite functions -

\lim_{x\rightarrow a}fog(x)=\lim_{x\rightarrow a}fg{(x)}
 

=\lim_{x\rightarrow b}f(x)
 

Where\:b=\lim_{x\rightarrow a}g(x)

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When x\rightarrow 1, then  \frac{\pi}{2}\sin\frac{\pi}{2} x \rightarrow \pi /2 \times 1 i.e \pi /2 

\therefore \lim_{x\rightarrow 1} \cos (\frac{\pi}{2}\sin\frac{\pi}{2} x)= \lim_{h\rightarrow \pi /2} \cos h=0   

(where h= \pi /2 \sin \pi x/2

 

 

 

 


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