lim-x-0-x-2-cos-x-3sin-x-x-5-

Question Number 131836 by liberty last updated on 09/Feb/21

\lim_{x \to 0} \frac{x (2 + \cos x) - 3 \sin x}{x^{5}}

Commented by EDWIN88 last updated on 10/Feb/21

another way L^{'} H \hat{o p i t a l}

\lim_{x \to 0} \frac{2 x + xcos x - 3 \sin x}{x^{5}} =

\lim_{x \to 0} \frac{2 + \cos x - xsin x - 3 \cos x}{{5 x}^{4}} = \lim_{x \to 0} \frac{2 - 2 \cos x - xsin x}{{5 x}^{4}}

\lim_{x \to 0} \frac{2 \sin x - (\sin x + xcos x)}{{20 x}^{3}} = \lim_{x \to 0} \frac{\sin x - xcos x}{{20 x}^{3}}

= \lim_{x \to 0} \frac{\cos x - (\cos x - xsin x)}{{60 x}^{2}}

= \lim_{x \to 0} \frac{xsin x}{{60 x}^{2}} = \frac{1}{60}

Answered by EDWIN88 last updated on 10/Feb/21

\lim_{x \to 0} \frac{x (2 + 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24}) - 3 (x - \frac{x^{3}}{6} + \frac{x^{5}}{120})}{x^{5}} =

\lim_{x \to 0} \frac{3 x - \frac{x^{3}}{2} + \frac{x^{5}}{24} - 3 x + \frac{x^{3}}{2} - \frac{x^{5}}{40}}{x^{5}} =

\lim_{x \to 0} \frac{x^{5} (\frac{1}{8.3} - \frac{1}{8.5})}{x^{5}} = \frac{5 - 3}{8.5.3} = \frac{1}{4.5.3} = \frac{1}{60}

Commented by malwan last updated on 10/Feb/21

t h a n k y o u s i r

Answered by malwan last updated on 09/Feb/21

\underset{x \to 0}{l i m} \frac{x (2 + 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} \mp . .) - 3 (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} \mp . .)}{x^{5}}

= \underset{x \to 0}{l i m} \frac{(3 x - \frac{x^{3}}{2} + \frac{x^{5}}{24} \mp . .) - (3 x - \frac{x^{3}}{2} + \frac{x^{5}}{40} \mp . .)}{x^{5}}

= \underset{x \to 0}{l i m} \frac{(\frac{5 - 3}{120}) x^{5} + . .}{x^{5}} = \frac{2}{120} = \frac{1}{60}

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