mathop {lim }limits_{x to 0} left[ {frac{1}{x | vedclass

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

Question

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} ight]$.Combine the terms

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} ight]$.Combine the terms into a single fraction:$\mathop {\lim }\limits_{x 	o 0} \left[ {\frac{x - \log (1 + x)}{{x^2}}} ight]$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'	ext{Hospital's rule}$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(x - \log (1 + x))}{\frac{d}{dx}(x^2)} = \mathop {\lim }\limits_{x 	o 0} \frac{1 - \frac{1}{1+x}}{2x}$.Simplify the expression:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{1+x-1}{1+x}}{2x} = \mathop {\lim }\limits_{x 	o 0} \frac{x}{2x(1+x)} = \mathop {\lim }\limits_{x 	o 0} \frac{1}{2(1+x)}$.Substitute $x = 0$:$\frac{1}{2(1+0)} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right] =$

Similar Questions

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

$\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8} = $

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module