Question

limx➡0 Tan3x-2x/3x-sin^2x

Answer 1

Hope it helepd you....

Answer 2

Answer:

$\lim_{x \to 0} \dfrac{\tan 3x-2x}{3x-\sin^2x}=\frac{1}{3}$

Step-by-step explanation:

Given : $\lim_{x \to 0} \dfrac{\tan 3x-2x}{3x-\sin^2x}$

Apply limits we get, $\frac{0}{0}$ form, thus,

Applying L' hopital rule,

Differentiating numerator and denominator with respect to x, we have,

$\lim_{x \to 0} \dfrac{3\sec^23x-2}{3-2\sin x\cosx}$

Also, $2\sin x \cos x =\sin2x$,

$\lim_{x \to 0} \dfrac{3\sec^23x-2}{3-\sin2x}$

Apply limits , we get,

$\Rightarrow \frac{3\sec^23(0)-2}{3-\sin2(0)}$

$\Rightarrow \frac{3(1)-2}{3}=\frac{1}{3}$

thus, $\lim_{x \to 0} \dfrac{\tan 3x-2x}{3x-\sin^2x}=\frac{1}{3}$