1-lim-x-0-tan-pi-4-x-cotx-2-lim-x-0-1-sin-x-1-x-

Question Number 167617 by mathlove last updated on 21/Mar/22

(1) \lim_{x \to 0} {[t a n (\frac{π}{4} - x)]}^{c o t x} = ?

(2) \lim_{x \to 0} [\frac{1}{\sin x} - \frac{1}{x}] = ?

Answered by cortano1 last updated on 21/Mar/22

(2) \lim_{x \to 0} (\frac{x - \sin x}{xsin x}) = \lim_{x \to 0} (\frac{x - \sin x}{x^{2} (\frac{\sin x}{x})})

= \lim_{x \to 0} (\frac{x - \sin x}{x^{2}}) = \lim_{x \to 0} (\frac{1 - \cos x}{2 x})

= 0

Answered by LEKOUMA last updated on 21/Mar/22

(2) \lim_{x \to 0} (\frac{1}{\sin x} - \frac{1}{x})

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

H o s p i t a l

\lim_{x \to 0} (\frac{{(x - \sin x)}^{'}}{{(x \sin x)}^{'}}) = \lim_{x \to 0} (\frac{{(x)}^{'} - {(\sin x)}^{'}}{{(x \sin x)}^{'}})

\lim_{x \to 0} \frac{1 - \cos x}{\sin x + x \cos x} = \lim_{x \to 0} \frac{\frac{1 - \cos x}{x}}{\frac{\sin x}{x} + \cos x} = \frac{0}{1 + 1} = \frac{0}{2} = 0

\lim_{x \to 0} (\frac{1}{\sin x} - \frac{1}{x}) = 0

Answered by qaz last updated on 21/Mar/22

\lim_{x \to 0} {[\tan (\frac{π}{4} - x)]}^{\cot x} = \lim_{x \to 0} {(\frac{\cot x - 1}{\cot x + 1})}^{\cot x} = \lim_{x \to 0} {[{(1 - \frac{2}{\cot x + 1})}^{\frac{\cot x + 1}{2}}]}^{2} {(1 - \frac{2}{\cot x + 1})}^{- 1} = e^{- 2}

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