Question

Evaluate of the following limits. $=\displaystyle \lim_{x \to 2}$ \left\\{\frac{\left(x^{2}-4\right)}{\sqrt{3x-2}-\sqrt{x+2}} \right\\}=

• A. $8$
• B. $16$
• C. $-5$
• D. $-8$

Answer 1

Answer: (A)

$8$

Answer 2

$=\displaystyle \lim_{x \to 2}$ \left\\{\frac{\left(x^{2}-4\right)}{\sqrt{3x-2}-\sqrt{x+2}} \right\\} $=\displaystyle \lim_{x \to 2}$ \left\\{\frac{\left(x^{2}-4\right)}{\sqrt{3x-2}-\sqrt{x+2}} \times \frac{\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{\left(\sqrt{3x-2}+\sqrt{x+2}\right)}\right\\} $=\displaystyle \lim_{x \to 2}$ \left\\{\frac{\left(x^{2}-4\right)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{\left(3x-2\right)-\left(x+2\right)} \right\\} $=\displaystyle \lim_{x \to 2}$ \left\\{\frac{\left(x^{2}-4\right)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2\left(x-2\right)} \right\\} $=\displaystyle \lim_{x \to 2}$ \left\\{\frac{\left(x+2\right)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2} \right\\}=8