Question

${lim}_{h\to 0}\{\frac{1}{h{(8+h)}^{1/3}}-\frac{1}{2h}\}$ is equal to:

• A. $\frac{1}{2}$
• B. $\frac{-4}{3}$
• C. $\frac{-16}{3}$
• D. $\frac{-1}{48}$

Answer 1

The correct option is D $\frac{-1}{48}$

$\underset{h\to 0}{lim}\left\{\frac{1}{h(8+h{)}^{1/3}}\right\}$ is equal to -

Solution $-\underset{h\to 0}{lim}\left\{\frac{2-(8+h{)}^{1/3}}{2h(8+h{)}^{1/3}}\right\}\frac{0}{0}$ form

$\Rightarrow \underset{h\to 0}{lim}\left\{\frac{-1/3(8+h{)}^{-2/3}}{2(8+h{)}^{1/3}+\frac{2}{3}h(8+h{)}^{-2/3}}\right\}$ D is hospital Rule

$\Rightarrow$ Applying limit, we get

$\Rightarrow \frac{\frac{-1}{3}\times (2^3{)}^{-2/3}}{2(2^3{)}^{1/3}+0}$

$\Rightarrow \frac{-1}{3}\times \frac{1}{4}\times \frac{1}{4}$

$\Rightarrow \frac{-1}{48}$

So, option (D) is right