Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs


$\huge \bf \color{navy}{QU} \color{purple}{ES} \color{blue}TI \color{skyblue}ON\ \textless \ br /\ \textgreater$

Find Thr Value of :-
$\bf \large \lim_{x \to 1 {}^{ - } } \frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } x} }{ \sqrt{1 - x} }$

➽ Explanation Is compulsory.
➽No Spam and No copy.
☆》 Otherwise Reported And Deleted​

Answer 1

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

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

To evaluate this limit,

$\rm :\longmapsto\:Put \: x = 1 - y \: \: so \: thay \: y \: \to \: 0$

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } (1 - y)} }{ \sqrt{1 - (1 - y)} }$

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } (1 - y)} }{ \sqrt{y} }$

If we directly Substitute y = 0, then

$\rm \: = \: \: \displaystyle \frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } (1 - 0)} }{ \sqrt{0} }$

$\rm \: = \: \: \displaystyle \frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } 1} }{ 0 }$

$\rm \: = \: \: \displaystyle \frac{ \sqrt{\pi} - \sqrt{2 \times \dfrac{\pi}{2} } }{ 0 }$

$\rm \: = \: \: \displaystyle \frac{ \sqrt{\pi} - \sqrt{ \pi} }{ 0 }$

$\rm \: = \: \: \displaystyle \frac{0 }{ 0 }$

which is meaningless.

So, Consider again

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } (1 - y)} }{ \sqrt{y} }$

Now Using L - Hospital Rule, we get

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \frac{\dfrac{d}{dy} \sqrt{\pi} -\dfrac{d}{dy} \sqrt{2 {sin}^{ - 1 } (1 - y)} }{\dfrac{d}{dy} \sqrt{y} }$

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \dfrac{0 - \dfrac{1}{2 \sqrt{ 2{sin}^{ - 1} (1 - y)}}\dfrac{d}{dy} 2{sin}^{ - 1}(1 - y)}{\dfrac{1}{2 \sqrt{y} } }$

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \dfrac{- 2 \: \dfrac{1}{\sqrt{2 {sin}^{ - 1} (1 - y)}}\dfrac{1}{ \sqrt{1 - {(1 - y)}^{2} } }\dfrac{d}{dy}(1 - y)}{\dfrac{1}{ \sqrt{y} } }$

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \dfrac{- 2 \: \dfrac{1}{\sqrt{ 2{sin}^{ - 1} (1 - y)}}\dfrac{1}{ \sqrt{1 - {(1 - y)}^{2} } }( - 1)}{\dfrac{1}{ \sqrt{y} } }$

$\rm \: = \: \: \displaystyle \lim_{y \to 0} \dfrac{ 2 \: \dfrac{1}{\sqrt{2 {sin}^{ - 1} (1 - y)}}\dfrac{1}{ \sqrt{1 - {(1 - y)}^{2} } }}{\dfrac{1}{ \sqrt{y} } }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{2 {sin}^{ - 1} 1} } \displaystyle\large \lim_{y \: \to 0} \frac{ \sqrt{y} }{ \sqrt{1 - {(1 - y)}^{2} } }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{2 \times \dfrac{\pi}{2}}} \displaystyle\large \lim_{y \: \to 0} \frac{ \sqrt{y} }{ \sqrt{1 - {(1 + y}^{2} - 2y) } }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{\pi}} \displaystyle\large \lim_{y \: \to 0} \frac{ \sqrt{y} }{ \sqrt{1 - {1 - y}^{2} + 2y} }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{\pi}} \displaystyle\large \lim_{y \: \to 0} \frac{ \sqrt{y} }{ \sqrt{2y- {y}^{2}} }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{\pi}} \displaystyle\large \lim_{y \: \to 0} \frac{ \sqrt{y} }{ \sqrt{y(2- {y})} }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{\pi}} \displaystyle\large \lim_{y \: \to 0} \frac{ 1 }{ \sqrt{(2- {y})} }$

$\rm \: = \: \:\dfrac{2}{ \sqrt{\pi}} \times \dfrac{1}{ \sqrt{2} }$

$\rm \: = \: \:\dfrac{ \sqrt{2} \times \sqrt{2} }{ \sqrt{\pi}} \times \dfrac{1}{ \sqrt{2} }$

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

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

Hence,

$\purple{ \boxed{\rm :\longmapsto \displaystyle \lim_{x \to 1 {}^{ - } } \bf\frac{ \sqrt{\pi} - \sqrt{2 {sin}^{ - 1 } x} }{ \sqrt{1 - x} } = \sqrt{\dfrac{2}{\pi} \: } \: \: \: \: \: \: \: \: }}$

Formula Used :-

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}x= 1}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx} {sin}^{ - 1}x = \frac{1}{\sqrt{1 - {x}^{2} } } }}}$

Additional Information :-

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

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

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

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

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

Answer 2

Answer:

$\underline{\purple{\ddot{\Mathsdude}}}$

♧♧Answered by Rohith kumar maths dude :-

♧♧I posted your answer on attachment refer it.

♤♤Hope it helps u @Parth yadav10056.