Find the limit.
-$\lim _ { x \rightarrow 1 ^ { + } } \left( \frac { 1 } { x ^ { 4 / 5 } } - \frac { 1 } { ( x - 1 ) ^ { 2 / 5 } } \right)$
A) $\approx$
B) Does not exist
C) $\infty$
D) 0

Question

Quizplus · Accepted Answer

To evaluate this limit, we need to examine the right-hand limit $\lim_{xightarrow 1^+}$ since $x$ approaches $1$ from the right. Let's evaluate the two terms separately.
$$\lim_{xightarrow 1^+} \frac{1}{x^{4/5}}=\frac{1}{1^{4/5}}=1$$
To evaluate the second term, we can use the following property of limits: 
If $\lim_{xightarrow c}(f(x)-g(x))=L$, then $\lim_{xightarrow c}g(x)=\lim_{xightarrow c}f(x)-L$. 
Applying this to the second term, we get
$$\lim_{xightarrow 1^+}\frac{1}{(x-1)^{2/5}}=\lim_{xightarrow 1^+}\left(\frac{1}{(x-1)^{2/5}} - \frac{1}{0^{2/5}}ight)=\lim_{xightarrow 1^+}\frac{1}{(x-1)^{2/5}}$$
The second equality follows because $1/0^\alpha$ ($\alpha>0$) is undefined, but as $xightarrow 1^+$, $(x-1)^{2/5}$ approaches $0$ and is positive, so we can replace $0^{2/5}$ by $0$.
Now we can evaluate this limit using L'Hopital's rule:
\begin{align*}
\lim_{xightarrow 1^+}\frac{1}{(x-1)^{2/5}} &= \lim_{xightarrow 1^+}\frac{d}{dx}\frac{1}{(x-1)^{2/5}}\
&= \lim_{xightarrow 1^+}\frac{2}{5(x-1)^{7/5}}\
&= \infty
\end{align*}
Therefore, the limit of the difference of the two terms is $1-\infty=-\infty$. Since the limit does not exist in the real numbers, the answer is (B) Does not exist. However, notice that the answer choices allow for approximation, and $\infty$ is a perfectly good approximation for $-\infty$ in this context. Therefore, the answer (A) $\approx \infty$ is also correct.

Find the Limit lim⁡x→1+(1x4/5−1(x−1)2/5)\lim _ { x \rightarrow 1 ^ { + } } \left( \frac { 1 } { x ^ { 4 / 5 } } - \frac { 1 } { ( x - 1 ) ^ { 2 / 5 } } \right)limx→1+​(x4/51​−(x−1)2/51​)

Find the Limit $\lim _ { x \rightarrow 1 ^ { + } } \left( \frac { 1 } { x ^ { 4 / 5 } } - \frac { 1 } { ( x - 1 ) ^ { 2 / 5 } } \right)$