Question

lim x=1 √4+x-√5 upon x-1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{4 + x} - \sqrt{5} }{x - 1}$

If we substitute directly x = 1, we get

$\rm \:  =  \: \dfrac{ \sqrt{4 + 1} - \sqrt{5} }{1 - 1}$

$\rm \:  =  \: \dfrac{ \sqrt{5} - \sqrt{5} }{0}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form

So, the above limit can be evaluated by using the Method of Rationalization

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{4 + x} - \sqrt{5} }{x - 1}$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{4 + x} - \sqrt{5} }{x - 1} \times \dfrac{ \sqrt{4 + x} + \sqrt{5} }{ \sqrt{4 + x} + \sqrt{5} }$

We know,

$\boxed{\tt{ (x - y)(x + y) = {x}^{2} - {y}^{2}}}$

So, using this, we get

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm \frac{4 + x - 5}{(x - 1)( \sqrt{4 + x} + \sqrt{5})}$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm \frac{x - 1}{(x - 1)( \sqrt{4 + x} + \sqrt{5})}$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm \frac{\cancel{x - 1}}{\cancel{(x - 1)} \: \: ( \sqrt{4 + x} + \sqrt{5})}$

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm \frac{1}{ \sqrt{4 + x} + \sqrt{5} }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{1 + 4} + \sqrt{5} }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{5} + \sqrt{5} }$

$\rm \:  =  \: \dfrac{1}{ 2\sqrt{5}}$

Hence,

$\rm \implies\:\:\boxed{\tt{ \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{4 + x} - \sqrt{5} }{x - 1} = \frac{1}{2 \sqrt{5} }}}$

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MORE TO KNOW

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