Calculate-lim-x-0-1-cos-1-cos-x-x-4-lim-x-0-1-cos-xcos-2xcos-3x-x-2-

Question Number 161409 by LEKOUMA last updated on 17/Dec/21

C a l c u l a t e

\lim_{x \to 0} \frac{1 - \cos (1 - \cos x)}{x^{4}}

\lim_{x \to 0} \frac{1 - \cos x \cos 2 x \cos 3 x}{x^{2}}

Commented by cortano last updated on 17/Dec/21

(2) \lim_{x \to 0} \frac{1 - \cos x \cos 2 x \cos 3 x}{x^{2}}

= \lim_{x \to 0} \frac{1 - \cos x + \cos x - \cos x \cos 2 x \cos 3 x}{x^{2}}

= \lim_{x \to 0} \frac{1 - \cos x}{x^{2}} + \lim_{x \to 0} \frac{\cos x (1 - \cos 2 x \cos 3 x)}{x^{2}}

= \frac{1}{2} + \lim_{x \to 0} \frac{1 - \cos 2 x + \cos 2 x - \cos 2 x \cos 3 x}{x^{2}}

= \frac{1}{2} + \lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}} + \lim_{x \to 0} \frac{\cos 2 x (1 - \cos 3 x)}{x^{2}}

= \frac{1}{2} + 2 + \lim_{x \to 0} \frac{1 - \cos 3 x}{x^{2}}

= \frac{5}{2} + \frac{9}{2} = \frac{14}{2} = 7

Answered by qaz last updated on 17/Dec/21

\lim_{x \to 0} \frac{1 - \cos xcos 2 xcos 3 x}{x^{2}}

= - \lim_{x \to 0} \frac{\ln (\cos xcos 2 xcos 3 x)}{x^{2}}

= - \lim_{x \to 0} \frac{lncos x}{x^{2}} - \lim_{x \to 0} \frac{lncos 2 x}{x^{2}} - \lim_{x \to 0} \frac{lncos 3 x}{x^{2}}

= \lim_{x \to 0} \frac{1 - \cos x}{x^{2}} + \lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}} + \lim_{x \to 0} \frac{1 - \cos 3 x}{x^{2}}

= \frac{1}{2} + \frac{1}{2} \cdot 2^{2} + \frac{1}{2} \cdot 3^{2}

= 7

- - - - - - - - - - - -

\lim_{x \to 0} \frac{1 - \cos (1 - \cos x)}{x^{4}}

= \lim_{x \to 0} \frac{\frac{1}{2} {(1 - \cos x)}^{2}}{x^{4}}

= \frac{1}{2} \lim_{x \to 0} \frac{{(\frac{1}{2} x^{2})}^{2}}{x^{4}}

= \frac{1}{8}

Commented by LEKOUMA last updated on 17/Dec/21

G o o d! M a n y t h a n k

Answered by cortano last updated on 17/Dec/21

(1) \lim_{x \to 0} \frac{1 - \cos (1 - \cos x)}{x^{4}}

= \lim_{x \to 0} \frac{2 \sin^{2} (\frac{1 - \cos x}{2})}{x^{4}}

= \lim_{x \to 0} \frac{2 \sin^{2} (\frac{2 \sin^{2} (\frac{x}{2})}{2})}{x^{4}}

= 2 \times \frac{4 \times \frac{1}{16}}{4} = 2 \times \frac{1}{16} = \frac{1}{8}

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