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\}=$

Updated On: Jul 6, 2022
  • $8$
  • $16$
  • $-5$
  • $-8$
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The Correct Option is A

Solution and Explanation

$=\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$
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