Question

solve X.
$\underset{x\to 0}{lim}\frac{5x\cos x+3\sin x}{3x^2+\tan x}$

Answer 1

$\underset{x\to 0}{lim}\frac{5x\cos x+3\sin x}{3x^2+\tan x}$ is of $\frac{0}{0}$ form

$=\underset{x\to 0}{lim}\frac{5\cos x-5x\sin x+3\cos x}{6x+{\sec }^{2}x}$

$=\underset{x\to 0}{lim}\frac{-5x\cos x+8\cos x}{6x+{\sec }^{2}x}$

$=\frac{0+8}{0+1}=8$