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Question Number 6939 by sou1618 last updated on 03/Aug/16

please solve L but do not use L′Hopital′s rule. L=lim_(x→0) ((1/x)−(1/(e^x −1)))

$$\mathrm{please}\:\mathrm{solve}\:{L} \\ $$$$\mathrm{but}\:\mathrm{do}\:\mathrm{not}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hopital}'\mathrm{s}\:\mathrm{rule}. \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\right) \\ $$

Commented by Yozzii last updated on 03/Aug/16

(1/x)−(1/(e^x −1))=((e^x −1−x)/(x(e^x −1)))=(((e^x −1−x)/x^2 )/((e^x /x)−(1/x))) e^x −1−x=(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!))+... ⇒((e^x −1−x)/x^2 )=(1/2)+(x/(3!))+(x^2 /(4!))+(x^3 /(5!))+... (e^x /x)−(1/x)=−(1/x)+(1/x)+1+(x/(2!))+(x^2 /(3!))+(x^3 /(4!))+... (e^x /x)−(1/x)=1+(x/(2!))+(x^2 /(3!))+(x^3 /(4!))+... ∴L=lim_(x→0) ((1/x)−(1/(e^x −1)))=lim_(x→0) (((1/2)+(x/(3!))+(x^2 /(4!))+(x^3 /(5!))+...)/(1+(x/(2!))+(x^2 /(3!))+(x^3 /(4!))+...)) L=(((1/2)+0+0+0+...)/(1+0+0+0+...))=(1/2)

$$\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{1}}=\frac{{e}^{{x}} −\mathrm{1}−{x}}{{x}\left({e}^{{x}} −\mathrm{1}\right)}=\frac{\frac{{e}^{{x}} −\mathrm{1}−{x}}{{x}^{\mathrm{2}} }}{\frac{{e}^{{x}} }{{x}}−\frac{\mathrm{1}}{{x}}} \\ $$$${e}^{{x}} −\mathrm{1}−{x}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+...\: \\ $$$$\Rightarrow\frac{{e}^{{x}} −\mathrm{1}−{x}}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{3}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{5}!}+...\: \\ $$$$\frac{{e}^{{x}} }{{x}}−\frac{\mathrm{1}}{{x}}=−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}}+\mathrm{1}+\frac{{x}}{\mathrm{2}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{4}!}+... \\ $$$$\frac{{e}^{{x}} }{{x}}−\frac{\mathrm{1}}{{x}}=\mathrm{1}+\frac{{x}}{\mathrm{2}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{4}!}+... \\ $$$$\therefore{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{3}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{5}!}+...}{\mathrm{1}+\frac{{x}}{\mathrm{2}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{4}!}+...} \\ $$$${L}=\frac{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{0}+\mathrm{0}+\mathrm{0}+...}{\mathrm{1}+\mathrm{0}+\mathrm{0}+\mathrm{0}+...}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Commented by sou1618 last updated on 03/Aug/16

Thank you very much

$${Thank}\:{you}\:{very}\:{much} \\ $$

Answered by Yozzii last updated on 03/Aug/16

Check answer in comments...

$${Check}\:{answer}\:{in}\:{comments}... \\ $$

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