Question

solve the following limit..​

Answer 1

Explanation :

$\bigstar \tt \: lim_{x \to0} \bigg( \cfrac{ \sqrt{1 + x} - \sqrt{1 - x} }{x} \bigg) \\ \\ \\ \underline{ \tt \:Applying \: L-Hospital \: rule, }\\ \\ \\ : \implies \tt lim_{x \to0} \bigg( \cfrac{ \cfrac{d}{dx}( \sqrt{1 + x} - \sqrt{1 - x} )}{ \cfrac{d}{dx} (x)} \bigg) \\ \\ \\ : \implies \tt lim_{x \to0} \bigg( \cfrac{ \cfrac{1}{2 \sqrt{1 + x} } + \cfrac{1}{2 \sqrt{1 - x} } }{1} \bigg) \\ \\ \\ : \implies \tt lim_{x \to0} \bigg( \cfrac{ \sqrt{1 - x} + \sqrt{1 + x} }{2( \sqrt{1 + x} \times \sqrt{1 + x} )} \bigg) \\ \\ \\ : \implies \tt lim_{x \to0} \bigg( \cfrac{ \sqrt{1 - x} + \sqrt{1 + x} } {2(1 - x {}^{2}) } \bigg) \\ \\ \\ : \implies \tt \bigg( \cfrac{ \sqrt{1 - 0} + \sqrt{1 + 0} }{2(1 - 0 {}^{2} )} \bigg) \\ \\ \\ : \implies \tt \bigg( \not{\cfrac{2}{2}} \bigg) \\ \\ \\ : \implies {\underline{ \boxed{ \tt1 }}} \: \bigstar$

Hence :

$\dag \boxed{\tt \: lim_{x \to0} \bigg( \cfrac{ \sqrt{1 + x} - \sqrt{1 - x} }{x} \bigg) = 1}$

Answer 2

Answer:

1

Step-by-step explanation:

By directly substituting the limits, we get (0/0) which is an indeterminate quantity. Therefore we have to choose different method.

Let's solve it by using method of rationalisation.

$\implies \lim \limits_{x \to 0} \dfrac{ \sqrt{1 + x} - \sqrt{1 - x} }{x}$

${ \implies \lim \limits_{x \to 0} \dfrac{ \sqrt{1 + x} - \sqrt{1 - x} }{x} \times \dfrac{ \sqrt{1 + x} + \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } }$

${ \implies \lim \limits_{x \to 0} \dfrac{ (\sqrt{1 + x} )^{2} - (\sqrt{1 - x})^{2} }{x(\sqrt{1 + x} + \sqrt{1 - x})}}$

${ \implies \lim \limits_{x \to 0} \dfrac{1 + x - 1 + x }{x(\sqrt{1 + x} + \sqrt{1 - x})}}$

${ \implies \lim \limits_{x \to 0} \dfrac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})}}$

${ \implies \lim \limits_{x \to 0} \dfrac{2}{\sqrt{1 + x} + \sqrt{1 - x}}}$

${ \implies \dfrac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}}}$

${ \implies \dfrac{2}{2}}$

$\implies 1$