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lim x->0 (cos(sinx)-cos(x))/x^4

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Note by Selena Miller
6 years, 9 months ago

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Comments

try the series expansion for sinx and cosx or u may also differentiate the numerator and denominator at least four times( l' Hospital's rule) to arrive at an answer .. !

Ramesh Goenka - 6 years, 9 months ago

want an easier method please And expansion didn't work

Selena Miller - 6 years, 9 months ago

For small xx, sin⁡(x)≈x−x36 \large \sin (x) \approx x - \frac {x^3}{6} , and cos⁡(x)≈1−x22 \large \cos (x) \approx 1 - \frac {x^2}{2}

lim⁡x→0cos⁡(sin⁡x)−cos⁡(x)x4=lim⁡x→0cos⁡(x−x36)−cos⁡(x)x4=lim⁡x→01−(x−x36)22−(1−x22)x4=lim⁡x→0x2−(x−x36)22x4=16 \begin{aligned} \LARGE \lim_{x \to 0} \frac { \cos (\sin x ) - \cos (x) } {x^4} & = & \lim_{x \to 0} \frac { \cos \left ( x - \frac {x^3}{6} \right ) - \cos (x) } {x^4} \\ \LARGE & = & \lim_{x \to 0} \frac { 1 - \frac { \left ( x - \frac {x^3}{6} \right )^2 }{2} - \left (1 - \frac {x^2}{2} \right ) }{x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {x^2 - \left ( x - \frac {x^3}{6} \right )^2 }{2x^4} = \boxed{ \frac {1}{6} } \\ \end{aligned}

Alternatively, recall that sum/difference to product identity cos⁡(A)−cos⁡(B)=−2sin⁡(A+B2)sin⁡(A−B2) \cos (A) - \cos (B) = -2 \sin \left ( \frac {A+B}{2} \right ) \sin \left ( \frac {A-B}{2} \right )

And this time, for small xx, sin⁡(x)≈x \sin (x) \approx x , apply L'hopital Rule

Let A=sin⁡(x),B=xA = \sin (x), B = x , we have

lim⁡x→0cos⁡(sin⁡x)−cos⁡(x)x4=lim⁡x→0−2sin⁡(sin⁡(x)+x2)sin⁡(sin⁡(x)−x2)x4=lim⁡x→0−2(sin⁡(x)+x2)(sin⁡(x)−x2)x4=lim⁡x→0sin⁡2(x)−x2−2x4=lim⁡x→0x2−sin⁡2(x)2x4 , Apply L’hopital Rule =lim⁡x→02x−2sin⁡(x)cos⁡(x)8x3=lim⁡x→02x−sin⁡(2x)8x3 , Apply L’hopital Rule again =lim⁡x→02−2cos⁡(2x)24x2 , Apply L’hopital Rule again =lim⁡x→04sin⁡(2x)24(2x)=16 \begin{aligned} \LARGE \lim_{x \to 0} \frac { \cos (\sin x ) - \cos (x) } {x^4} & = & \lim_{x \to 0} \frac {-2 \sin \left ( \frac {\sin (x) + x}{2} \right ) \sin \left ( \frac {\sin (x) - x}{2} \right ) }{x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {-2 \left ( \frac {\sin (x) + x}{2} \right ) \left ( \frac {\sin (x) - x}{2} \right ) }{x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {\sin^2 (x) - x^2 }{-2x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {x^2 - \sin^2 (x) }{2x^4} \space \text{, Apply L'hopital Rule } \\ \LARGE & = & \lim_{x \to 0} \frac {2x - 2 \sin (x) \cos(x) }{8x^3} \\ \LARGE & = & \lim_{x \to 0} \frac {2x - \sin (2x) }{8x^3} \space \text{, Apply L'hopital Rule again } \\ \LARGE & = & \lim_{x \to 0} \frac {2 - 2\cos (2x) }{24x^2} \space \text{, Apply L'hopital Rule again } \\ \LARGE & = & \lim_{x \to 0} \frac {4\sin (2x) }{24(2x)} = \boxed{ \frac {1}{6} } \\ \end{aligned}

Pi Han Goh - 6 years, 9 months ago
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