Question

$\underset{x\to 0}{lim}\frac{x^2}{1-\sqrt[4]{41+x^2}}=$

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

Answer 1

Answer: (C)

$-4$

$\underset{x\to 0}{lim}\frac{x^2}{1-\sqrt[4]{41+x^2}}$

Let $(1+x^2)=t^4$
So, $x\to 0$
$t\to 1$

$\underset{t\to 1}{lim}\frac{t^4-1}{1-t}$
$=\underset{t\to 1}{lim}\frac{t^4-1}{-(t-1)}$
$=\underset{t\to 1}{lim}\frac{(t^2-1)(t^2+1)}{-(t-1)}$
$=\underset{t\to 1}{lim}\frac{(t-1)(t+1)(t^2+1)}{-(t-1)}$
$=\underset{t\to 1}{lim}\frac{-(t+1)(t^2+1)}{1}$
$=-(1+1)\times (1^2+1)=-4$