lim _{x rightarrow 0} frac{2 sin x-sin 2 x}{x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $L = \lim _{x \rightarrow 0} \frac{2 \sin x - \sin 2x}{x^3}$ (which is of the form $\frac{0}{0}$).Applying $L$-Hospital rule:$L = \lim _{x \ri

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $L = \lim _{x \rightarrow 0} \frac{2 \sin x - \sin 2x}{x^3}$ (which is of the form $\frac{0}{0}$).Applying $L$-Hospital rule:$L = \lim _{x \rightarrow 0} \frac{2 \cos x - 2 \cos 2x}{3x^2}$ (which is of the form $\frac{0}{0}$).Applying $L$-Hospital rule again:$L = \lim _{x \rightarrow 0} \frac{-2 \sin x + 4 \sin 2x}{6x}$ (which is of the form $\frac{0}{0}$).Applying $L$-Hospital rule again:$L = \lim _{x \rightarrow 0} \frac{-2 \cos x + 8 \cos 2x}{6}$.Substituting $x = 0$:$L = \frac{-2 \cos(0) + 8 \cos(0)}{6} = \frac{-2(1) + 8(1)}{6} = \frac{6}{6} = 1$.

$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to

Similar Questions

Evaluate the limit: $\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$

$\lim _{x \rightarrow 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ is

Let $a > 0$ be a real number. Then the limit $\lim _{x \rightarrow 2} \frac{a^x+a^{3-x}-\left(a^2+a\right)}{a^{3-x}-a^{x / 2}}$ is

$\mathop {\lim }\limits_{\theta \to \pi /2} (\sec \theta - \tan \theta ) = $

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module