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प्रश्न
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
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उत्तर
\[\lim_{x \to 4} \left[ \frac{x^2 - 16}{\sqrt{x} - 2} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 4} \left[ \frac{x^2 - 4^2}{\sqrt{x} - 2} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left( x - 4 \right)\left( x + 4 \right)}{\left( \sqrt{x} - 2 \right)} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left\{ \left( \sqrt{x} \right)^2 - 2^2 \right\}\left( x + 4 \right)}{\left( \sqrt{x} - 2 \right)} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left( \sqrt{x} - 2 \right)\left( \sqrt{x} + 2 \right)\left( x + 4 \right)}{\left( \sqrt{x} - 2 \right)} \right]\]
\[ = \left( 2 + 2 \right)\left( 4 + 4 \right)\]
\[ = 32\]
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