Answers

2014-01-21T12:13:19+08:00
First calculus question I've seen haha yay:)

 \lim_{x \to 0} \frac{sin4x}{sin6x}
substituting x with zero you get \frac{0}{0} therefore you can apply l'hospital's rule and take the derivative of the numerator and the denominator
\lim_{x \to 0} \frac{sin4x}{sin6x} = \lim_{x \to 0}  \frac{ \frac{d}{dx} sin4x }{ \frac{d}{dx} sin6x}
= \lim_{x \to 0}  \frac{4cos4x}{6cos6x}
=   \frac{4cos4(0)}{6cos6(0)}
=  \frac{2}{3}
0