Question

$\underset{h\to 0}{lim}\{\frac{1}{h^3\sqrt{8+h}}-\frac{1}{2h}\}=$

• A. $\frac{-1}{12}$
• B. $\frac{-4}{3}$
• C. $\frac{-16}{3}$
• D. $\frac{-1}{48}$

Answer 1

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

$\underset{h\to 0}{lim}[\frac{1}{h(8+h{)}^1/3}-\frac{1}{2h}]$

$\frac{0}{0}$ form, so applying L- hospital rule,

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

$\Rightarrow \underset{h\to 0}{lim}\left[\frac{-1/3(8{)}^2/3}{2(8{)}^1/3+0}\right]=\left[\frac{-1}{6\left(8\right)}\right]=\frac{1}{48}$

So option (d) is correct.