lim _{x rightarrow 0} frac{left(1+frac{x}{2}r | vedclass

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

Question

(C) To evaluate the limit $\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x}$,we observe that it is in the indeterminate form $\f

Vedclass · Accepted Answer

$\frac{5}{14}$

Vedclass · Answer

(C) To evaluate the limit $\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x}$,we observe that it is in the indeterminate form $\frac{0}{0}$.We can apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}\left(\left(1+\frac{x}{2}\right)^{5 / 7}-1\right)}{\frac{d}{dx}(x)}$$= \lim _{x}$ ${\rightarrow 0} \frac{\frac{5}{7}\left(1+\frac{x}{2}\right)^{5 / 7 - 1} \cdot \frac{d}{dx}\left(1+\frac{x}{2}\right)}{1}$$= \lim _{x \rightarrow 0} \frac{5}{7}\left(1+\frac{x}{2}\right)^{-2 / 7} \cdot \frac{1}{2}$$= \frac{5}{7} \cdot (1+0)^{-2 / 7} \cdot \frac{1}{2}$$= \frac{5}{14}$Thus,the correct option is $C$.

$\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x} = $

Similar Questions

If $f(9) = 9$ and $f'(9) = 4$,then $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}} = $

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

Evaluate the given limit: $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}$

If $f$ is a real function such that $f(4)=4$ and $f^{\prime}(4)=16$,then $\lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2} =$

Let $a > 0$ be a real number. Then the limit $\lim _{x \rightarrow 2} \frac{a^x+a^{3-x}-\left(a^2+a\right)}{a^{3-x}-a^{x / 2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module