Question

please help me to solve this problem
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Answer 1

GIVEN :–

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$\: \: \: \bf \displaystyle \bf \: \lim_{x \to0} \frac{ \sqrt{a + x} - \sqrt{a} }{x}$

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TO FIND :–

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▪︎ Value = ?

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SOLUTION :–

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• Let the function –

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$\implies \bf \:A \: = \displaystyle \bf \: \lim_{x \to0} \frac{ \sqrt{a + x} - \sqrt{a} }{x}$

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• Using rationalization method –

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$\bf \implies A= \displaystyle \bf \: \lim_{x \to0}\left[ \dfrac{ \sqrt{a + x} - \sqrt{a} }{x} \times \dfrac{ \sqrt{a + x} + \sqrt{a}}{ \sqrt{a + x} + \sqrt{a}} \right]$

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$\bf \implies A= \displaystyle \bf \: \lim_{x \to0}\left[ \dfrac{( \sqrt{a + x} - \sqrt{a} )(\sqrt{a + x} + \sqrt{a})}{x( \sqrt{a + x} + \sqrt{a})} \right]$

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• Using identity –

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$\: \blacktriangleright \: \bf (a + b)(a - b) = {a}^{2} - {b}^{2}$

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• So that –

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$\implies \bf \:A \: = \displaystyle \bf \: \lim_{x \to0}\left[ \dfrac{( \sqrt{a + x})^{2} - (\sqrt{a})^{2}}{x( \sqrt{a + x} + \sqrt{a})} \right]$

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$\implies \bf \:A \: = \displaystyle \bf \: \lim_{x \to0}\left[ \dfrac{(a + x) - (a)}{x( \sqrt{a + x} + \sqrt{a})} \right]$

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$\implies \bf \:A \: = \displaystyle \bf \: \lim_{x \to0}\left[ \dfrac{ \cancel x}{ \cancel x( \sqrt{a + x} + \sqrt{a})} \right]$

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$\implies \bf \:A \: = \displaystyle \bf \: \lim_{x \to0}\left[ \dfrac{1}{( \sqrt{a + x} + \sqrt{a})} \right]$

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• Now Apply limits –

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$\implies \bf \:A \: = \left[ \dfrac{1}{( \sqrt{a +(0)} + \sqrt{a})} \right]$

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$\implies \bf \:A \: = \dfrac{1}{ \sqrt{a} + \sqrt{a} }$

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$\implies \large{ \boxed{ \bf \:A \: = \dfrac{1}{ 2\sqrt{a}}}}$