Math, asked by Tandeep7422, 11 months ago

Evaluate lim x~pi/2 (tan2x/x -pi/2)

Answers

Answered by Sharad001
60

QuesTion :-

 \rm Evaluate  \:  \:  \: \lim_{x \to  \frac{ \pi}{2} } \frac{ \tan2x}{x -   \frac{ \pi}{2} } \:  \\

Answer :-

 \mapsto \:  \boxed{ \rm \lim_{x \to  \frac{ \pi}{2} } \frac{ \tan2x}{x -   \frac{ \pi}{2} } \:   = 2} \:  \\

Used Limit property :-

 \implies \boxed{ \rm \: \lim_{} \frac{ \tan \: h }{h} \:  = 1} \\

Solution :-

We have ,

\mapsto \:  \rm \: \lim_{x \to  \frac{ \pi}{2} } \frac{ \tan2x}{x -   \frac{ \pi}{2} } \:  \\  \:  \\  \rm \: let \: t = x -  \frac{ \pi}{2}  \\ \rm hence \\  \\  \to \rm t +  \frac{ \pi}{2}  = x \\  \\ \rm replace \: these \: in \: given \: limit \\  \\  \mapsto \: \: \lim_{x \to  \frac{ \pi}{2} } \frac{ \tan2x}{x -   \frac{ \pi}{2} } \:  \\  \\ \rm \: after \: replacing \: these  \\  \\  \mapsto \rm \lim_{t \to 0 } \frac{ \tan2( \frac{ \pi}{2} + t)} {t } \:  \\  \:  \\  \because \rm \: x \to \:  \frac{ \pi}{2}  \\  \therefore \rm \: t\to \:  \frac{ \pi}{2}  -  \frac{ \pi}{2}  \to \: 0 \\  \\  \mapsto \rm \:  \lim_{t \to 0 } \frac{ \tan( { \pi} + 2t)} {t } \:  \\ \\  \sf multiply \: and \: divide \: by \: 2 \\  \\  \mapsto \: \rm \lim_{t \to 0 } \: 2 \frac{ \tan( { \pi} + 2t)} {2t } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:  \:  \:  \:  \:  \boxed{  \because \:  \: \rm \:  \tan( \pi +  \theta) =  \tan \theta } \\ \\  \mapsto \: \rm \lim_{t \to 0 } \:  \: 2 \frac{ \tan( 2t)} {2t } \\ \:  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \because \: \boxed{ \rm \: \lim_{} \frac{ \tan \: h }{h} \:  = 1} \\ \:  \\  \mapsto \: \rm \lim_{t \to 0 } \:  \: 2  \:  \times 1 \\  \\ \sf taking \: limit \:  \\  \\  \mapsto \: 2 \:  \\  \\ \:  \mapsto \:  \boxed{ \rm \lim_{x \to  \frac{ \pi}{2} } \frac{ \tan2x}{x -   \frac{ \pi}{2} } \:   = 2}

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