Question

Find b

$\underset{x\to 0}{lim}\frac{\tan 2x-x}{3x-\sin x}.=\frac{1}{b}.$

Answer 1

$\underset{x\to 0}{lim}\frac{\tan 2x-x}{3x-\sin x}\left[Form\frac{0}{0}\right]$
So using L-Hospital's rule,
$=\underset{x\to 0}{lim}\frac{\frac{d}{dx}(\tan 2x-x)}{\frac{d}{dx}(3x-\sin x)}$
$=\underset{x\to 0}{lim}\frac{2{\sec }^{2}2x-1}{3-\cos x}=\frac{2-1}{3-1}=\frac{1}{2}$
$\Rightarrow b=2$