mathop {lim }limits_{x to 0} frac{{{{(1 + x)} | vedclass

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

Question

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x}$,we can use the rationalization method o

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x}$,we can use the rationalization method or $L$-Hospital's rule.Method $1$: RationalizationMultiply the numerator and denominator by the conjugate $\frac{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}}{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}}$:$\mathop {\lim }\limits_{x 	o 0} \frac{{(1 + x) - (1 - x)}}{{x({{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}})}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2x}}{{x({{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}})}} = \mathop {\lim }\limits_{x 	o 0} \frac{2}{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}} = \frac{2}{{1 + 1}} = 1$.Method $2$: $L$-Hospital's ruleSince the limit is in the form $\frac{0}{0}$,differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}})}}{{\frac{d}{{dx}}(x)}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{2}{{(1 + x)}^{ - 1/2}} - ( - \frac{1}{2}{{(1 - x)}^{ - 1/2}})}}{1} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }} ight) = \frac{1}{2} + \frac{1}{2} = 1$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x} = $

Similar Questions

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 $f(x)$ is a differentiable function of $x$,then $\mathop {Limit}\limits_{h \to 0} \frac{f(x + 3h) - f(x - 2h)}{h} = $

$\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

If $f(3) = 6$ and $f'(3) = 2$,then $\mathop {\text{Limit}}\limits_{x \to 3} \frac{x f(3) - 3 f(x)}{x - 3}$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module