Question

Evaluate: $\underset{h\to 0}{lim}(\frac{1}{h\sqrt[3]{38+h}}-\frac{1}{2h})$

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

Answer 1

Answer: (D)

$-\frac{1}{48}$

Let $L=\underset{h\to 0}{lim}(\frac{1}{h\sqrt[3]{38+h}}-\frac{1}{2h})=\underset{h\to 0}{lim}\frac{1}{h}(\frac{1}{\sqrt[3]{38+h}}-\frac{1}{2})$

Let $a=\sqrt[3]{38+h}\Rightarrow a^3=8+h$ and $b=2\Rightarrow b^3=8$

Also $a^3-b^3=(a-b)(a^2+b^2+ab)=8+h-8=h\Rightarrow$ (1)

$\therefore L=\underset{h\to 0}{lim}\frac{1}{h}(\frac{1}{a}-\frac{1}{b})=-\underset{h\to 0}{lim}\frac{a-b}{h}\cdot \frac{1}{ab}$

$=-\underset{h\to 0}{lim}\frac{a^3-b^3}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$

$=-\underset{h\to 0}{lim}\frac{h}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$ .....using $\left(1\right)$

As $h\to 0\Rightarrow a^3=8\Rightarrow a=2$

$=-\frac{1}{2\times 2(2^2+2^2+2\times 2)}=$

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