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Question Number 77856 by walidMTS last updated on 11/Jan/20

l_(x→0) im(((e^x −x−1)/x))

$$\underset{{x}\rightarrow\mathrm{0}} {{l}im}\left(\frac{{e}^{{x}} −{x}−\mathrm{1}}{{x}}\right) \\ $$

Commented by mathmax by abdo last updated on 11/Jan/20

we have e^x =Σ_(n=0) ^∞ (x^n /(n!)) =1+x+(x^2 /(2!)) +(x^3 /(3!))+... ⇒ e^x =1+x+(x^2 /2) +o(x^3 ) (x→0) ⇒e^x −1−x =(x^2 /2) +o(x^3 ) ⇒ ((e^x −x−1)/x)=(x/2) +o(x^2 ) ⇒lim_(x→0) ((e^x −x−1)/x) =lim_(x→0) (x/2)=0

$${we}\:{have}\:{e}^{{x}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\:=\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+...\:\Rightarrow \\ $$$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\:\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{e}^{{x}} −\mathrm{1}−{x}\:=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$$\frac{{e}^{{x}} −{x}−\mathrm{1}}{{x}}=\frac{{x}}{\mathrm{2}}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{e}^{{x}} −{x}−\mathrm{1}}{{x}}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}}{\mathrm{2}}=\mathrm{0} \\ $$

Answered by Rio Michael last updated on 11/Jan/20

lim_(x→0) (((e^x −x−1)/x)) = ((lim_(x→0) (e^x −x−1))/(lim_(x→0) x)) = lim_(x→0) e^x −1 = 0

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{e}^{{x}} −{x}−\mathrm{1}}{{x}}\right)\:=\:\frac{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({e}^{{x}} −{x}−\mathrm{1}\right)}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{e}^{{x}} −\mathrm{1}\:=\:\mathrm{0} \\ $$

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