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Question Number 86655 by jagoll last updated on 30/Mar/20

lim_(x→−1^+ ) (((√π) −(√(arc cos x)))/(√(x+1)))

$$\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\pi}\:−\sqrt{\mathrm{arc}\:\mathrm{cos}\:\mathrm{x}}}{\sqrt{\mathrm{x}+\mathrm{1}}} \\ $$

Answered by john santu last updated on 30/Mar/20

L′hopital rule lim_(x→−1^+ ) ((−(1/2)(arc cos x)^(−(1/2)) (((−1)/(√(1−x^2 )))))/(1/(2(√(x+1))))) lim_(x→−1^+ ) (1/(√(arc cos x))) ×((√(1+x))/(√(1−x^2 ))) lim_(x→−1^+ ) (1/(√(arc cos x)))×(1/(√(1−x))) (1/(√π)) × (1/(√2)) = (1/(√(2π)))

$$\mathrm{L}'\mathrm{hopital}\:\mathrm{rule} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arc}\:\mathrm{cos}\:\mathrm{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{−\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\right)}{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}}}\: \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{\sqrt{\mathrm{arc}\:\mathrm{cos}\:\mathrm{x}}}\:×\frac{\sqrt{\mathrm{1}+\mathrm{x}}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{\sqrt{\mathrm{arc}\:\mathrm{cos}\:\mathrm{x}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\pi}}\:×\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}\pi}} \\ $$

Commented by jagoll last updated on 30/Mar/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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