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Question Number 29460 by prof Abdo imad last updated on 08/Feb/18
find lim_(x→0) ((e^(√(1+sinx)) −e)/(tanx)).
$${find}\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{e}^{\sqrt{\mathrm{1}+{sinx}}} \:\:−{e}}{{tanx}}. \\ $$
Answered by Cheyboy last updated on 09/Feb/18
lim_(x→0) ((e^(√(1+sin x)) −e)/(tanx))=0
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}} −{e}}{\mathrm{tan}{x}}=\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
for x−→0 we have 1+sinx ∼ 1 +x and (√(1+x)) ∼ 1+(x/2) and e^(1+(x/2)) ∼e(1+(x/2)) ⇒ e^(√(1+sinx)) −e ∼ ((ex)/2) but tanx∼x ⇒ lim_(x→0) ((e^(√(1+sinx)) −e)/(tanx)) = (e/2) .
$${for}\:{x}−\rightarrow\mathrm{0}\:\:{we}\:{have}\:\:\mathrm{1}+{sinx}\:\sim\:\mathrm{1}\:+{x}\:\:{and} \\ $$$$\sqrt{\mathrm{1}+{x}}\:\sim\:\mathrm{1}+\frac{{x}}{\mathrm{2}}\:\:{and}\:{e}^{\mathrm{1}+\frac{{x}}{\mathrm{2}}} \sim{e}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${e}^{\sqrt{\mathrm{1}+{sinx}}} \:−{e}\:\sim\:\frac{{ex}}{\mathrm{2}}\:\:{but}\:\:{tanx}\sim{x}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{e}^{\sqrt{\mathrm{1}+{sinx}}} −{e}}{{tanx}}\:=\:\frac{{e}}{\mathrm{2}}\:. \\ $$
Answered by sma3l2996 last updated on 09/Feb/18
sinx∼_0 x⇔1+sinx∼_0 1+x (√(1+sinx))∼_0 (1+x)^(1/2) (√(1+x))∼_0 1+(1/2)x e^(√(1+sinx)) ∼_0 e^(1+(1/2)x) =e×e^(x/2) e^(x/2) ∼_0 1+(x/4) e^(√(1+sinx)) −e∼_0 e(1+(x/4))−e=(e/4)x tanx∼_0 x lim_(x→0) ((e^(√(1+sinx)) −e)/(tanx))∼lim_(x→0) (((e/4)x)/x)=(e/4)
$${sinx}\underset{\mathrm{0}} {\sim}{x}\Leftrightarrow\mathrm{1}+{sinx}\underset{\mathrm{0}} {\sim}\mathrm{1}+{x} \\ $$$$\sqrt{\mathrm{1}+{sinx}}\underset{\mathrm{0}} {\sim}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}+{x}}\underset{\mathrm{0}} {\sim}\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$${e}^{\sqrt{\mathrm{1}+{sinx}}} \underset{\mathrm{0}} {\sim}{e}^{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}} ={e}×{e}^{\frac{{x}}{\mathrm{2}}} \\ $$$${e}^{{x}/\mathrm{2}} \underset{\mathrm{0}} {\sim}\mathrm{1}+\frac{{x}}{\mathrm{4}} \\ $$$${e}^{\sqrt{\mathrm{1}+{sinx}}} −{e}\underset{\mathrm{0}} {\sim}{e}\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)−{e}=\frac{{e}}{\mathrm{4}}{x} \\ $$$${tanx}\underset{\mathrm{0}} {\sim}{x} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{\sqrt{\mathrm{1}+{sinx}}} −{e}}{{tanx}}\sim\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{{e}}{\mathrm{4}}{x}}{{x}}=\frac{{e}}{\mathrm{4}} \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
for u ∈v(0) e^u ∼ 1+u so e^(x/2) ∼1+(x/2) .
$$\:{for}\:{u}\:\in{v}\left(\mathrm{0}\right)\:\:{e}^{{u}} \:\sim\:\mathrm{1}+{u}\:{so}\:{e}^{\frac{{x}}{\mathrm{2}}} \sim\mathrm{1}+\frac{{x}}{\mathrm{2}}\:\:. \\ $$
Commented by sma3l2996 last updated on 22/Feb/18
Yeah, you′re right
$${Yeah},\:{you}'{re}\:{right} \\ $$

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