## Calculus: Early Transcendentals (2nd Edition)

$\lim_{x\to0}\dfrac{\sin3x}{\tan4x}=\dfrac{3}{4}$
$\lim_{x\to0}\dfrac{\sin3x}{\tan4x}$ Substitute $\tan4x$ by $\dfrac{\sin4x}{\cos4x}$: $\lim_{x\to0}\dfrac{\sin3x}{\tan4x}=\lim_{x\to0}\dfrac{\sin3x}{\Big(\dfrac{\sin4x}{\cos4x}\Big)}=...$ $...=\lim_{x\to0}\dfrac{\sin3x\cos4x}{\sin4x}=...$ Divide the numerator and the denominator by $(3x)(4x)$: $...=\lim_{x\to0}\dfrac{\Big(\dfrac{1}{4x}\Big)\Big(\dfrac{\sin3x}{3x}\Big)\cos4x}{\Big(\dfrac{\sin4x}{4x}\Big)\Big(\dfrac{1}{3x}\Big)}=...$ Simplify: $...=\lim_{x\to0}\dfrac{3x\Big(\dfrac{\sin3x}{3x}\Big)\cos4x}{4x\Big(\dfrac{\sin4x}{4x}\Big)}=\dfrac{3}{4}\lim_{x\to0}\dfrac{\Big(\dfrac{\sin3x}{3x}\Big)\cos4x}{\Big(\dfrac{\sin4x}{4x}\Big)}=...$ Use the product and quotient limit laws to evaluate the limit: $...=\Big(\dfrac{3}{4}\Big)\dfrac{\lim_{x\to0}\Big(\dfrac{\sin3x}{3x}\Big)\cos4x}{\lim_{x\to0}\dfrac{\sin4x}{4x}}=...$ $...=\Big(\dfrac{3}{4}\Big)\dfrac{\lim_{x\to0}\Big(\dfrac{\sin3x}{3x}\Big)\lim_{x\to0}\cos4x}{\lim_{x\to0}\dfrac{\sin4x}{4x}}=...$ Evaluate the limits and simplify: $...=\Big(\dfrac{3}{4}\Big)\Big[\dfrac{(1)(1)}{1}\Big]=\dfrac{3}{4}$