Provide an appropriate response.
-Evaluate $\lim _ { x \rightarrow 1 } \frac { x ^ { 2 / 7 } - 1 } { x + 1 } =$ by first converting it to a derivative at a particular $x$-value.
A) $- \frac { 1 } { 7 }$
B) $- \frac { 7 } { 2 }$
C) $- \frac { 2 } { 7 }$
D) $\frac { 2 } { 7 }$

Question

Quizplus · Accepted Answer

To convert the limit to a derivative, we use L'Hôpital's Rule:

\begin{align*}
\lim_{xightarrow 1} \frac{x^{2/7}-1}{x+1} &= \lim_{xightarrow 1} \frac{\frac{2}{7}x^{-5/7}}{1} \
&= \frac{2}{7}\left(\frac{1}{1}ight)^{-5/7} \
&= -\frac{2}{7} \
\end{align*}

Therefore, the answer is C.

Provide an Appropriate Response lim⁡x→1x2/7−1x+1=\lim _ { x \rightarrow 1 } \frac { x ^ { 2 / 7 } - 1 } { x + 1 } =limx→1​x+1x2/7−1​=

Provide an Appropriate Response $\lim _ { x \rightarrow 1 } \frac { x ^ { 2 / 7 } - 1 } { x + 1 } =$