×

# help

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

Note by Selena Miller
2 years, 10 months ago

Sort by:

For small $$x$$, $$\large \sin (x) \approx x - \frac {x^3}{6}$$, and $$\large \cos (x) \approx 1 - \frac {x^2}{2}$$

$\begin{eqnarray} \LARGE \lim_{x \to 0} \frac { \cos (\sin x ) - \cos (x) } {x^4} & = & \lim_{x \to 0} \frac { \cos \left ( x - \frac {x^3}{6} \right ) - \cos (x) } {x^4} \\ \LARGE & = & \lim_{x \to 0} \frac { 1 - \frac { \left ( x - \frac {x^3}{6} \right )^2 }{2} - \left (1 - \frac {x^2}{2} \right ) }{x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {x^2 - \left ( x - \frac {x^3}{6} \right )^2 }{2x^4} = \boxed{ \frac {1}{6} } \\ \end{eqnarray}$

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

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

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

$\begin{eqnarray} \LARGE \lim_{x \to 0} \frac { \cos (\sin x ) - \cos (x) } {x^4} & = & \lim_{x \to 0} \frac {-2 \sin \left ( \frac {\sin (x) + x}{2} \right ) \sin \left ( \frac {\sin (x) - x}{2} \right ) }{x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {-2 \left ( \frac {\sin (x) + x}{2} \right ) \left ( \frac {\sin (x) - x}{2} \right ) }{x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {\sin^2 (x) - x^2 }{-2x^4} \\ \LARGE & = & \lim_{x \to 0} \frac {x^2 - \sin^2 (x) }{2x^4} \space \text{, Apply L'hopital Rule } \\ \LARGE & = & \lim_{x \to 0} \frac {2x - 2 \sin (x) \cos(x) }{8x^3} \\ \LARGE & = & \lim_{x \to 0} \frac {2x - \sin (2x) }{8x^3} \space \text{, Apply L'hopital Rule again } \\ \LARGE & = & \lim_{x \to 0} \frac {2 - 2\cos (2x) }{24x^2} \space \text{, Apply L'hopital Rule again } \\ \LARGE & = & \lim_{x \to 0} \frac {4\sin (2x) }{24(2x)} = \boxed{ \frac {1}{6} } \\ \end{eqnarray}$ · 2 years, 10 months ago

want an easier method please And expansion didn't work · 2 years, 10 months ago