mathop {lim }limits_{x to 1} frac{{1 + log x | vedclass

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

Question

(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 1$.Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 1$.Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(1 + \log x - x)}{\frac{d}{dx}(1 - 2x + x^2)} = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{1}{x} - 1}{-2 + 2x}$$= \mathop {\lim }\limits_{x 	o 1} \frac{1 - x}{x(-2 + 2x)} = \mathop {\lim }\limits_{x 	o 1} \frac{-(x - 1)}{2x(x - 1)}$$= \mathop {\lim }\limits_{x 	o 1} \frac{-1}{2x} = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \log x - x}}{{1 - 2x + {x^2}}} = $

Similar Questions

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

$\lim\limits_{x \rightarrow 0} \frac{\int\limits_{0}^{x} t \sin (10 t) d t}{x}$ is equal to

$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $

If $f(4) = g(4) = 2$,$f'(4) = 9$,and $g'(4) = 6$,then $\mathop {\text{Limit}}\limits_{x \to 4} \frac{\sqrt{f(x)} - \sqrt{g(x)}}{\sqrt{x} - 2}$ is equal to:

If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,then as $x$ approaches $a$,the limit of $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module