Question

please help ma frnds answer is 1​

Answer 1

Answer:

answer is 1

hope it helps it

Answer 2

Appropriate Question

Evaluate :-

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \rm \dfrac{x}{ \sqrt{1 + x} - \sqrt{1 - x} }$

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

The given expression is

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \rm \dfrac{x}{ \sqrt{1 + x} - \sqrt{1 - x} } }$

If we substitute x = 0, directly we get

$\rm \: = \: \: \dfrac{0}{ \sqrt{1 + 0} - \sqrt{1 - 0} }$

$\rm \: = \: \: \dfrac{0}{ \sqrt{1} - \sqrt{1} }$

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

So, it means to evaluate

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \rm \dfrac{x}{ \sqrt{1 + x} - \sqrt{1 - x} } }$

we use Method of Rationalization.

So,

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \rm \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}} \times \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}$

We know,

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

So, we get

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \rm \frac{x(\sqrt{1 + x} + \sqrt{1 - x})}{ {( \sqrt{x + 1} )}^{2} - {( \sqrt{1 - x}) }^{2} }$

$\rm \: = \: \:\displaystyle\lim_{x \to 0} \rm \dfrac{x(\sqrt{1 + x} + \sqrt{1 - x})}{x + 1 - (1 - x)}$

$\rm \: = \: \:\displaystyle\lim_{x \to 0} \rm \dfrac{x(\sqrt{1 + x} + \sqrt{1 - x})}{x + 1 - 1 + x}$

$\rm \: = \: \:\displaystyle\lim_{x \to 0} \rm \dfrac{x(\sqrt{1 + x} + \sqrt{1 - x})}{2x}$

$\rm \: = \: \:\displaystyle\lim_{x \to 0} \rm \dfrac{(\sqrt{1 + x} + \sqrt{1 - x})}{2}$

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

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

$\rm \: = \: \: \dfrac{1 + 1}{2}$

$\rm \: = \: \: \dfrac{2}{2}$

$\rm \: = \: \: 1$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \bf \dfrac{x}{ \sqrt{1 + x} - \sqrt{1 - x} } \: = \: 1$

Additional Information :-

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

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

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

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

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