Math, asked by alaganikavya, 26 days ago

please help ma frnds answer is 1​

Attachments:

Answers

Answered by sharmadsatyam
0

Answer:

answer is 1

hope it helps it

Answered by mathdude500
1

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}}

Similar questions