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

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(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} = $

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