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Evaluate $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2}  – \sqrt {x + 2} }}$.
Evaluate $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2}  – \sqrt {x + 2} }}$.
CBSE | Class 11 | Exercise 1 | Limits and Derivatives | Maths | NCERT Exemplar
Evaluate $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2}  – \sqrt {x + 2} }}$.

We are given, $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2}  – \sqrt {x + 2} }}$

Rationalize the denominator to get,

$ = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2}  – \sqrt {x + 2} }} \times \left( {\frac{{\sqrt {3x – 2}  + \sqrt {x + 2} }}{{\sqrt {3x – 2}  + \sqrt {x + 2} }}} \right)$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} – 4} \right)(\sqrt {3x – 2}  + \sqrt {x + 2} )}}{{(3x – 2) – (x + 2)}}$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{(x – 2)(x + 2)(\sqrt {3x – 2}  + \sqrt {x + 2} )}}{{2x – 4}}$

Grouping the like terms and simplifying to get,

$ = \mathop {\lim }\limits_{x \to 2} \frac{{(x – 2)(x + 2)(\sqrt {3x – 2}  + \sqrt {x + 2} )}}{{2(x – 2)}}$

Substituting the limits to get,

$ = \mathop {\lim }\limits_{x \to 2} \frac{{(x + 2)(\sqrt {3x – 2}  + \sqrt {x + 2} )}}{2}$

Therefore, $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2}  – \sqrt {x + 2} }} = 8$.

 

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