Question

$\underset{x\to 0}{lim}\frac{x^2}{1-\sqrt[3]{31+x^2}}$=

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

Answer 1

The correct option is C $-3$

We know, $\underset{x\to 0}{lim}(1+x{)}^n\approx (1+nx)$
$\therefore \underset{x\to 0}{lim}\frac{x^2}{1-\sqrt[3]{31+x^2}}=\underset{x\to 0}{lim}\frac{x^2}{1-(1+x^2{)}^{1/3}}$
$=\underset{x\to 0}{lim}\frac{x^2}{1-(1+\frac{1}{3}{x}^2)}=\underset{x\to 0}{lim}\frac{x^2}{-\frac{1}{3}{x}^2}=-3$