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प्रश्न
If f (1) = 1, f' (1) = 2, then write the value of \[\lim_{x \to 1} \frac{\sqrt{f (x)} - 1}{\sqrt{x} - 1}\]
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उत्तर
\[\lim_{x \to 1} \frac{\sqrt{f\left( x \right)} - 1}{\sqrt{x} - 1}\]
\[ = \lim_{x \to 1} \frac{\sqrt{f\left( x \right)} - 1}{\sqrt{x} - 1} \times \frac{\sqrt{f\left( x \right)} + 1}{\sqrt{f\left( x \right)} + 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1}\]
\[ = \lim_{x \to 1} \frac{\left( f\left( x \right) - 1 \right)\left( \sqrt{x} + 1 \right)}{\left( x - 1 \right)\left( \sqrt{f\left( x \right)} + 1 \right)}\]
\[ = \lim_{x \to 1} \frac{f\left( x \right) - 1}{x - 1} \times \lim_{x \to 1} \frac{\left( \sqrt{x} + 1 \right)}{\left( \sqrt{f\left( x \right)} + 1 \right)}\]
\[ = f'\left( 1 \right) \times \frac{1 + 1}{\sqrt{f\left( 1 \right)} + 1}\]
\[ = 2 \times \frac{2}{1 + 1}\]
\[ = 2\]
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