Lim_x→ 1√(π) √(cos^ 1x)/√(x+1) Indeterminate form Using L' hospital's rule lim_x→ 10 1/2√(cos^ 1x) ( 1/√(1 x^2) )/1/2√(x+1)(1+0) lim_x→ 11/√(1 x^2)√(cos^ 1x) ...

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Byju's Answer

Standard XII

Mathematics

Existence of Limit

lim x →-1√π-√...

Question

${lim}_{x \to - 1} \frac{\sqrt{π} - \sqrt{c o s^{- 1} x}}{\sqrt{x + 1}}$

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Solution

${lim}_{x \to - 1} \frac{\sqrt{π} - \sqrt{c o s^{- 1} x}}{\sqrt{x + 1}}$
Indeterminate form
Using L' hospital's rule
${lim}_{x \to - 1} \frac{0 - \frac{1}{} 2 \sqrt{c o s^{- 1} x} (\frac{- 1}{\sqrt{1 - x^{2}}})}{\frac{1}{2 \sqrt{x + 1}} (1 + 0)}$
${lim}_{x \to 1} \frac{\frac{1}{\sqrt{1 - x^{2}} \sqrt{c o s^{- 1} x}}}{\frac{1}{\sqrt{x + 1}}}$
$= {lim}_{x \to - 1} \frac{\sqrt{x + 1}}{\sqrt{1 - x^{2}} \sqrt{c o s^{- 1} x}} = {lim}_{x \to - 1} \frac{1}{\sqrt{1 - x} \sqrt{c o s^{- 1} x}}$
$= \frac{1}{\sqrt{1 (- 1)} \sqrt{c o s^{- 1} (- 1)}} = \frac{1}{\sqrt{2} \sqrt{π}}$
$= \frac{1}{\sqrt{π}}$

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Similar questions

lim x \to - 1 \frac{\sqrt{π} - \sqrt{{cos}^{- 1} x}}{\sqrt{x + 1}}

lim x \to - 1 \frac{\sqrt{π} - \sqrt{{cos}^{- 1} x}}{\sqrt{x + 1}} =

Q. Solve

lim x \to - 1^{+} \frac{\sqrt{π} - \sqrt{{cos}^{- 1} x}}{\sqrt{x + 1}} =

lim x \to 1^{+} \frac{\sqrt{π} \sqrt{{cos}^{- 1} x}}{\sqrt{x + 1}}

is equal to

lim x \to - 1^{+} \frac{\sqrt{π} - \sqrt{{cos}^{- 1} x}}{\sqrt{x + 1}}

is equal to

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limx→−1√π−√cos−1x√x+1

limx→−1√π−√cos−1x√x+1 Indeterminate form Using L' hospital's rule limx→−10−12√cos−1x(−1√1−x2)12√x+1(1+0) limx→11√1−x2√cos−1x1√x+1 =limx→−1√x+1√1−x2√cos−1x=limx→−11√1−x√cos−1x =1√1(−1)√cos−1(−1)=1√2√π =1√π

${lim}_{x \to - 1} \frac{\sqrt{π} - \sqrt{c o s^{- 1} x}}{\sqrt{x + 1}}$