Question

Evaluate : $\underset{x\to 0}{\lim }\frac{\sqrt{1+x+x^2}-1}{x}$

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

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\frac{\sqrt{1+x+x^2}-1}{x}$
$=\underset{x\to 0}{\lim }\frac{\sqrt{1+x+x^2}-1}{x}.\frac{\sqrt{1+x+x^2}+1}{\sqrt{1+x+x^2}+1}$
$=\underset{x\to 0}{\lim }\frac{1+x+x^2-1}{x\left(\sqrt{1+x+x^2}+1\right)}$
$=\underset{x\to 0}{\lim }\frac{x\left(1+x\right)}{x\left(\sqrt{1+x+x^2}+1\right)}$
$=\underset{x\to 0}{\lim }\frac{1+x}{\sqrt{1+x+x^2}+1}=\frac{1}{2}$