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Question Number 115162 by john santu last updated on 24/Sep/20

 (1)  lim_(x→1)  arc sin (((1−(√x))/(1−x))) =?  (2) lim_(x→∞)  e^(x^3 +(√(cos ((1/x^2 )))))  =?  (3) lim_(x→0)  csc x .sin (sin x) =?

$$\:\left(\mathrm{1}\right)\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}\right)\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{x}^{\mathrm{3}} +\sqrt{\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \:=? \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{csc}\:{x}\:.\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\:=? \\ $$

Answered by bobhans last updated on 24/Sep/20

(1) lim_(x→1)  arc sin (((1−(√x))/(1−x))) = arc sin (lim_(x→1) ((1−(√x))/(1−x)))                 = arc sin (lim_(x→1)  ((1−(√x))/((1−(√x))(1+(√x)))))                = arc sin (lim_(x→1)  (1/(1+(√x))))               = arc sin ((1/2)) = (π/6)

$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}\right)\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{{x}}}{\left(\mathrm{1}−\sqrt{{x}}\right)\left(\mathrm{1}+\sqrt{{x}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{x}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{6}} \\ $$

Answered by bobhans last updated on 24/Sep/20

(2) lim_(x→∞)  e^(x^3  +(√(cos ((1/x^2 )))))  = e^(lim_(x→∞)  x^3 +(√(cos ((1/x^2 )))))    = e^∞  = ∞

$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{x}^{\mathrm{3}} \:+\sqrt{\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \:=\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} +\sqrt{\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \\ $$$$\:=\:{e}^{\infty} \:=\:\infty\: \\ $$

Answered by bobhans last updated on 24/Sep/20

(3) lim_(x→0)  csc x. sin (sin x) = lim_(x→0)  ((sin (sin x))/(sin x))  [ let sin x = w ; w→0 ]  lim_(x→0)  ((sin (sin x))/(sin x)) = lim_(w→0)  ((sin (w))/w) = 1

$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{csc}\:{x}.\:\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:{x}} \\ $$$$\left[\:{let}\:\mathrm{sin}\:{x}\:=\:{w}\:;\:{w}\rightarrow\mathrm{0}\:\right] \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:{x}}\:=\:\underset{{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({w}\right)}{{w}}\:=\:\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 24/Sep/20

lim_(x→1) sin^(−1) (((1−(√x))/(1+x)))=sin^(−1) ((1/(1+(√x))))=(π/6)

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\mathrm{1}+\mathrm{x}}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{x}}}\right)=\frac{\pi}{\mathrm{6}} \\ $$

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