The value of mathop {lim }limits_{x to 0} fra | vedclass

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

Question

(D) We are given the limit $\mathop {\lim }\limits_{x 	o 0} \frac{2}{x}\log (1 + x)$.Using the property of logarithms,$a \log b = \log b^a$,we can re

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We are given the limit $\mathop {\lim }\limits_{x 	o 0} \frac{2}{x}\log (1 + x)$.Using the property of logarithms,$a \log b = \log b^a$,we can rewrite the expression as $\mathop {\lim }\limits_{x 	o 0} 2 \log (1 + x)^{\frac{1}{x}}$.Since the logarithm function is continuous,we have $2 \log \left( \mathop {\lim }\limits_{x 	o 0} (1 + x)^{\frac{1}{x}} ight)$.We know that $\mathop {\lim }\limits_{x 	o 0} (1 + x)^{\frac{1}{x}} = e$.Therefore,the expression becomes $2 \log_e e = 2 	imes 1 = 2$.Alternatively,using $L'Hospital's$ rule: $\mathop {\lim }\limits_{x 	o 0} \frac{2 \log(1+x)}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{2 \cdot \frac{1}{1+x}}{1} = \frac{2}{1} = 2$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x)$ is equal to

Similar Questions

$\lim _{x \rightarrow 2} \frac{x+3 x^2+5 x^3+7 x^4-166}{x-2} = $

If $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$,$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \pi/2} \frac{\cos x}{x - \pi/2}$,then:

If $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2,$ then $a + b + c$ is equal to ...........

$\mathop {\lim }\limits_{x \to 0^+} {x^x} = $

If $\lim _{x}$ ${\rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}}$,where $\operatorname{gcd}(m, n)=1$,then $8 m+12 n$ is equal to.

Vedclass Test Series

Exam Paper Generator

Online Exam Module