Question

a1" title="\lim_{x \to 0 } \frac{x-sin(2x)}{x+sin(3x)}" alt="\lim_{x \to 0 } \frac{x-sin(2x)}{x+sin(3x)}" align="absmiddle" class="latex-formula">

Answer 1

Step-by-step explanation:

$= \lim \limits_{x \to0} \frac{x - \sin(2x) }{x + \sin(3x) }$

$= \lim \limits_{x \to0} \frac{ \frac{d}{dx}(x - \sin(2x) ) }{ \frac{d}{dx} (x + \sin(3x) )}$

$= \lim \limits_{x \to0} \frac{1 - 2\cos(2x) }{1 + 3 \cos(3x) }$

$= \lim \limits_{x \to0} \frac{ \frac{d}{dx}(1 + 2 \cos(2x) ) }{ \frac{d}{dx} (1 + 3 \cos(3x) )}$

$= \lim \limits_{x \to0} \frac{ - 4 \sin(2x) }{ - 9 \sin(3x) }$

$= \frac{ - 4.2}{ - 9.3}$

$= \frac{ - 8}{ - 27}$

$= \frac{8}{27}$