lim-x-0-x-tan-pi-2-1-x-

Question Number 203247 by mathlove last updated on 13/Jan/24

\lim_{x \to 0} x t a n \frac{π}{2} (1 + x) = ?

Answered by MM42 last updated on 13/Jan/24

= {l i m}_{x \to 0} - x c o t \frac{π}{2} x = {l i m}_{x \to 0} - \frac{\frac{π}{2} x c o s \frac{π}{2} x}{s i n \frac{π}{2} x} \times \frac{2}{π}

= - \frac{2}{π} ✓

Commented by mathlove last updated on 13/Jan/24

w a y t a n \frac{π}{2} (1 + x) = - c o t x

Commented by MM42 last updated on 13/Jan/24

o k

- c o t \frac{π}{2} x

Answered by mr W last updated on 13/Jan/24

\lim_{x \to 0} x \tan \frac{π}{2} (1 + x)

= \lim_{x \to 0} \frac{x \sin (\frac{π}{2} + \frac{π x}{2})}{\cos (\frac{π}{2} + \frac{π x}{2})}

= \lim_{x \to 0} \frac{x \sin (\frac{π}{2} - \frac{π x}{2})}{- \cos (\frac{π}{2} - \frac{π x}{2})}

= \lim_{x \to 0} \frac{x \cos (\frac{π x}{2})}{- \sin (\frac{π x}{2})}

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

= - \frac{2}{π}

Commented by mathlove last updated on 13/Jan/24

t h a t s r i g h t

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