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

Question

Aicharya · Accepted Answer

$\lim _{x ightarrow 0} \frac{\left(1+\frac{x}{2}ight)^{5 / 7}-1}{x}$ Here, Limit is $\div$ form, so we can apply L-Hospital rule $ \begin{aligned} & =\lim _{x ightarrow 0} \frac{\frac{5}{7}\left(1+\frac{x}{2}ight)^{-2 / 7} \cdot \frac{d}{d x}\left(\frac{x}{2}ight)-0}{1} \ & =\lim _{x ightarrow 0} \frac{5}{7}\left(1+\frac{x}{2}ight)^{-2 / 7} \cdot \frac{1}{2} \ & =\frac{5}{7}(1+0)^{-2 / 7} 	imes \frac{1}{2}=\frac{5}{14} \end{aligned} $ Hence, option (3) is correct

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

Solution