Question

can anybody please prove that​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Prove that

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \frac{ \sqrt{x} - 1 + \sqrt{x - 1} }{ \sqrt{ {x}^{2} - 1} } = \frac{1}{ \sqrt{2} }$

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

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \frac{ \sqrt{x} - 1 + \sqrt{x - 1} }{ \sqrt{ {x}^{2} - 1} }$

If we put directly x = 1, we get

$\rm \:  =  \:\dfrac{ \sqrt{1} - 1 - \sqrt{1 - 1} }{ \sqrt{1 - 1} }$

$\rm \:  =  \:\dfrac{1 - 1 - 0 }{ 0}$

$\rm \:  =  \:\dfrac{0}{ 0} \: which \: is \: meaningless$

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \frac{ \sqrt{x} - 1 + \sqrt{x - 1} }{ \sqrt{ {x}^{2} - 1} }$

can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{x \to 1} \frac{ \sqrt{x} - 1 }{ \sqrt{ {x}^{2} - 1} } + \displaystyle\lim_{x \to 1} \frac{ \sqrt{x - 1} }{ \sqrt{ {x}^{2} - 1} }$

$\rm \:  =  \:\displaystyle\lim_{x \to 1} \frac{ \sqrt{x} - 1 }{ \sqrt{ {x}^{2} - 1} } \times \frac{ \sqrt{x} + 1}{ \sqrt{x} + 1} + \displaystyle\lim_{x \to 1} \frac{ \sqrt{x - 1} }{ \sqrt{(x - 1)(x + 1)} }$

We know,

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

So, using this identity, we get

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

$\rm \:  =  \:\displaystyle\lim_{x \to 1} \frac{ {( \sqrt{x - 1} )}^{2} }{ \sqrt{(x - 1)(x + 1)}( \sqrt{x} + 1)} + \displaystyle\lim_{x \to 1} \frac{1}{ \sqrt{x + 1} }$

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

$\rm \:  =  \:\dfrac{ \sqrt{1 - 1} }{ \sqrt{1 + 1} ( \sqrt{1} + 1)} + \dfrac{1}{ \sqrt{2} }$

$\rm \:  =  \:\dfrac{0 }{ \sqrt{2} ( 1 + 1)} + \dfrac{1}{ \sqrt{2} }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{2} }$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \frac{ \sqrt{x} - 1 + \sqrt{x - 1} }{ \sqrt{ {x}^{2} - 1} } = \frac{1}{ \sqrt{2} }$

Additional Information :-

$\boxed{ \rm \:\displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1}$

$\boxed{ \rm \:\displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1}$

$\boxed{ \rm \:\displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1}$

$\boxed{ \rm \:\displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1}$

$\boxed{ \rm \:\displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}$