Question

Evaluate of the following limits.
$\displaystyle \lim_{x \to 0}$$\frac{\left(\sqrt{1+3x}-\sqrt{1-3x}\right)}{x}=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$\displaystyle \lim_{x \to 0}$$\frac{\left(\sqrt{1+3x}-\sqrt{1-3x}\right)}{x}$
$=\displaystyle \lim_{x \to 0}$$\frac{\left(\sqrt{1+3x}-\sqrt{1-3x}\right)}{x}$ $\times \frac{\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}{\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}$
$=\displaystyle \lim_{x \to 0}$$\frac{\left\{\left(1+3x\right)-\left(1-3x\right)\right\}}{x\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}$
$=\displaystyle \lim_{x \to 0}$$\frac{6x}{x\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}$
$=\displaystyle \lim_{x \to 0}$$\frac{6}{\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}$
$=\frac{6}{\left(\sqrt{1}+\sqrt{1}\right)}=\frac{6}{2}=3$.