Question

$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+x}-\sqrt[3]{31-x}}{x}$=

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $0$
• D. $1$

Answer 1

Answer: (B)

$\frac{2}{3}$

$L=\underset{x\to 0}{lim}\frac{\sqrt[3]{31+x}-\sqrt[3]{31-x}}{x}$
$L=\underset{x\to 0}{lim}\frac{{(x+1)}^{1/3}-{(1-x)}^{1/3}}{x}$
$=\underset{x\to 0}{lim}\frac{{\left[\right(x+1)}^{1/3}-1]{-\left[\right(1-x)}^{1/3}-1]}{x}$
$=\underset{x\to 0}{lim}\frac{{(x+1)}^{1/3}-1}{x}-\underset{x\to 0}{lim}\frac{{(1-x)}^{1/3}-1}{x}$
$=\underset{x\to 0}{lim}\frac{{(x+1)}^{1/3}-1}{x+1-1}+\underset{x\to 0}{lim}\frac{{(1-x)}^{1/3}-1}{1-x-1}$
Since, $x$ tends to $0$
$\Rightarrow 1+x$ tends to $1$ and $1-x$ also tends to $1$
So $L=\frac{1}{3}+\frac{1}{3}$ ($\because {lim}_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$)
$=\frac{2}{3}$