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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 22 of 72
Marks: +1, -0
lim⁡x→π/2tan⁡2xx−π/2\lim\limits_{x\to \pi/2} \frac{\tan 2x}{x-\pi/2}
Solution:  
We have, lim⁡x→π/2tan⁡2xx−π/2\lim\limits_{x\to \pi/2} \frac{\tan 2x}{x-\pi/2} (0/0 form)
Put x = π2\frac{\pi}{2} + y; if x → π2\frac{\pi}{2} ⇒ y → 0
lim⁡y→0tan⁡2(π2+y)π2+y−π2\lim\limits_{y\to 0} \frac{\tan 2\left(\frac{\pi}{2}+y\right)}{\frac{\pi}{2}+y-\frac{\pi}{2}} = lim⁡y→0tan⁡(π+2y)y\lim\limits_{y\to 0} \frac{\tan(\pi+2y)}{y} = (lim⁡x→0tan⁡2y2y)\left( \lim\limits_{x\to 0} \frac{\tan 2y}{2y} \right) × 2 = 2.
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