mathop {lim }limits_{x to 0} frac{{sin x + lo | 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{{\sin x + \lo

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(C) Using $L$-Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\cos x - \frac{1}{{1 - x}}}}{{2x}}$Applying $L$-Hospital's rule again:$= \mathop {\lim }\limits_{x 	o 0} \frac{{ - \sin x - \frac{1}{{{{(1 - x)}^2}}}}}{2} = \frac{{ - 0 - 1}}{2} = - \frac{1}{2}$.Alternatively,using Taylor series expansions:$\sin x = x - \frac{{{x^3}}}{{3!}} + \dots$$\log (1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \dots$Substituting these into the limit:$= \mathop {\lim }\limits_{x 	o 0} \frac{{(x - \frac{{{x^3}}}{6} + \dots) + (- x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \dots)}}{{{x^2}}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{ - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{2} - \dots}}{{{x^2}}} = - \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}}$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to \pi /2} \frac{2x - \pi}{\cos x} = $

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

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

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

Let $f(1) = g(1) = k$ and their $n^{th}$ derivatives $f^{(n)}(1), g^{(n)}(1)$ exist and are not equal for some $n$. If $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module