mathop {lim }limits_{x to 0} frac{{log _e}(1 | vedclass

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

Question

(D) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log _e}(1 + x)}{{3^x - 1}}$This is in the $\frac{0}{0}$ indeterminate form.Applying $L'	ex

Vedclass · Accepted Answer

$\log _3 e$

Vedclass · Answer

(D) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log _e}(1 + x)}{{3^x - 1}}$This is in the $\frac{0}{0}$ indeterminate form.Applying $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}[\log _e(1 + x)]}{\frac{d}{dx}[3^x - 1]} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{1}{1 + x}}{3^x \log _e 3}$Substituting $x = 0$:$= \frac{\frac{1}{1 + 0}}{3^0 \log _e 3} = \frac{1}{1 \cdot \log _e 3} = \frac{1}{\log _e 3}$Using the property $\frac{1}{\log _a b} = \log _b a$,we get:$= \log _3 e$.

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

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$ is equal to

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

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:

The value of $\lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$ is

$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module