mathop {lim }limits_{x to 0} frac{{x({e^x} - | vedclass

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

Question

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}}$.Using the standard limits $\mathop {\lim }\l

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}}$.Using the standard limits $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - 1}}{x} = 1$ and the identity $1 - \cos x = 2 \sin^2(x/2)$:$\mathop {\lim }\limits_{x 	o 0} \frac{{x({e^x} - 1)}}{{2 \sin^2(x/2)}}$.Multiply and divide by $x$ and $(x/2)^2$ to use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{{{e^x} - 1}}{x} ight) \cdot \frac{x^2}{2 \sin^2(x/2)}$.$= 1 \cdot \mathop {\lim }\limits_{x 	o 0} \frac{x^2}{2 (x/2)^2 \cdot (\frac{\sin(x/2)}{x/2})^2}$.$= \mathop {\lim }\limits_{x 	o 0} \frac{x^2}{2 (x^2/4) \cdot 1^2} = \frac{1}{2/4} = 2$.

$\mathop {\lim }\limits_{x \to 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}} = $

Similar Questions

If $\lim_{x \to \infty} \frac{(2x - 1)^{19} \cdot (3x + 2)^{11}}{(6x - 5)^{30}} = 2^a \cdot 3^b$, then $a + b = $

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x),$ where $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$

$\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=$

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module