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Question Number 160501 by LEKOUMA last updated on 30/Nov/21

Calculate  1) lim_(x→1) ((cos ((Π/2))x)/(1−(√x)))  2) lim_(x→+∞) (e^(1+x) /((1+x)^x ))−(x/e)

$${Calculate} \\ $$$$\left.\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{2}}\right){x}}{\mathrm{1}−\sqrt{{x}}} \\ $$$$\left.\mathrm{2}\right)\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{e}^{\mathrm{1}+{x}} }{\left(\mathrm{1}+{x}\right)^{{x}} }−\frac{{x}}{{e}} \\ $$

Commented by cortano last updated on 01/Dec/21

(1) lim_(x→1)  ((cos ((π/2)x))/(1−(√x)))    let 1−(√x) = u ; x=(1−u)^2    lim_(u→0)  ((cos ((π/2)(1−u)^2 ))/u)   = lim_(u→0)  ((sin (π/2)(u^2 −2u))/u)   =lim_(u→0)  (((π/2)u(u−2))/u) = −π

$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{1}−\sqrt{\mathrm{x}}}\: \\ $$$$\:\mathrm{let}\:\mathrm{1}−\sqrt{\mathrm{x}}\:=\:\mathrm{u}\:;\:\mathrm{x}=\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{2}} \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{2}} \right)}{\mathrm{u}} \\ $$$$\:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{2u}\right)}{\mathrm{u}} \\ $$$$\:=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\pi}{\mathrm{2}}\mathrm{u}\left(\mathrm{u}−\mathrm{2}\right)}{\mathrm{u}}\:=\:−\pi \\ $$

Answered by qaz last updated on 01/Dec/21

lim_(x→1) ((cos (π/2)x)/(1−(√x)))=lim_(x→1) ((cos (π/2)x)/(−ln(√x)))=lim_(x→1) ((−(π/2)sin (π/2)x)/(−(1/(2x))))=π

$$\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\mathrm{x}}{\mathrm{1}−\sqrt{\mathrm{x}}}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\mathrm{x}}{−\mathrm{ln}\sqrt{\mathrm{x}}}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{−\frac{\pi}{\mathrm{2}}\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\mathrm{x}}{−\frac{\mathrm{1}}{\mathrm{2x}}}=\pi \\ $$

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