lim-x-1-x-1-tan-pix-2-

Question Number 119074 by benjo_mathlover last updated on 22/Oct/20

\lim_{x \to 1} (x - 1) \tan (\frac{π x}{2})

Answered by Dwaipayan Shikari last updated on 22/Oct/20

\lim_{x \to 1} (x - 1) c o t (\frac{π}{2} - \frac{π}{2} x)

= \lim_{x \to 1} - (1 - x) \frac{c o s (1 - x) \frac{π}{2}}{s i n (1 - x) \frac{π}{2}}

= \lim_{x \to 1} - (1 - x) \frac{1}{(1 - x) \frac{π}{2}} = - \frac{2}{π} \lim_{x \to 0} s i n x = x

Answered by bemath last updated on 22/Oct/20

l e t x = 1 + m \land m \to 0

\lim_{m \to 0} \frac{m}{\cot (\frac{(1 + m) π}{2})} = \lim_{m \to 0} [\frac{m}{- \tan (\frac{m π}{2})}]

= - \frac{2}{π}

Answered by 1549442205PVT last updated on 22/Oct/20

Put x - 1 = t \Rightarrow (x \to 1 \sim t \to 0 \sim π t / 2 \to 0)

I = \lim (x - 1) \tan (\frac{π x}{2}) = \lim_{t \to 0} [ttan \frac{π (1 + t)}{2}]

= \lim_{t \to 0} (- tcot \frac{π t}{2}) = \lim_{t \to 0} (- \frac{1}{π}) (\frac{π t}{\sin \frac{π t}{2}} \times \cos \frac{π t}{2})

= \lim_{t \to 0} (- \frac{1}{π}) \times \frac{2}{\frac{\sin (π t / 2)}{(π t / 2)}} \times \cos (π t / 2) =

- \frac{2}{π} \times 1 = - \frac{2}{π}

Answered by malwan last updated on 22/Oct/20

\underset{x \to 1}{l i m} (x - 1) c o t (\frac{π}{2} - \frac{π x}{2})

= \underset{x \to 1}{l i m} (x - 1) c o t \frac{π}{2} (1 - x)

= \underset{x \to 1}{l i m} \frac{\frac{π}{2} (1 - x) \times (- 1)}{\frac{π}{2} t a n \frac{π}{2} (1 - x)} = - \frac{2}{π}

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