mathop {lim }limits_{x to 0} frac{{tan 2x - x | vedclass

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

Question

(C) Using $L-Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{	an 2x - x}

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(C) Using $L-Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{	an 2x - x}}{{3x - \sin x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}(	an 2x - x)}}{{\frac{d}{{dx}}(3x - \sin x)}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{2\sec^2 2x - 1}}{{3 - \cos x}}$Substituting $x = 0$:$= \frac{{2\sec^2(0) - 1}}{{3 - \cos(0)}} = \frac{{2(1)^2 - 1}}{{3 - 1}} = \frac{{2 - 1}}{2} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\log _e}(1 + x)}{{3^x - 1}} = $

Let $f(x)$ be a differentiable function and $f^{\prime}(4)=5$. Then, $\lim _{x \rightarrow 2} \frac{f(4) - f\left(x^{2}\right)}{x-2}$ equals

$\mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$ equals

$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$ is equal to

If $f^{\prime \prime}(0)=k, k \neq 0,$ then the value of $\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module