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Question Number 160569 by LEKOUMA last updated on 02/Dec/21

Calculate   1.  lim_(x→0) [2e^(x/(x+1)) −1]^((x^2 +1)/x)   2. lim_(x→0) (((1+x×2^x )/(1+x×3^x )))^(1/x^2 )

$${Calculate}\: \\ $$$$\mathrm{1}.\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{2}{e}^{\frac{{x}}{{x}+\mathrm{1}}} −\mathrm{1}\right]^{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}} \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{x}×\mathrm{2}^{{x}} }{\mathrm{1}+{x}×\mathrm{3}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$ \\ $$

Answered by qaz last updated on 02/Dec/21

(1)::lim_(x→0) [2e^(x/(x+1)) −1]^((x^2 +1)/x)   =lim_(x→0) [2(1+(x/(x+1))+o((x/(x+1))))−1]^((x^2 +1)/x)   =lim_(x→0) [(1+((2x)/(x+1))+o((x/(x+1))))^((x+1)/(2x)) ]^((2(x^2 +1))/(x+1))   =e^2   (2)::lim_(x→0) (((1+2^x x)/(1+3^x x)))^(1/x^2 )   =elim_(x→0) ((ln(1+2^x x)−ln(1+3^x x))/x^2 )  =elim_(x→0,ξ→1) (1/(ξx^2 ))[(1+2^x x)−(1+3^x x)]  =elim_(x→0) ((2^x −3^x )/x)  =elim_(x→0) (2^x ln2−3^x ln3)  =(2/3)

$$\left(\mathrm{1}\right)::\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{2e}^{\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}} −\mathrm{1}\right]^{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}+\mathrm{o}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)\right)−\mathrm{1}\right]^{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{2x}}{\mathrm{x}+\mathrm{1}}+\mathrm{o}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)\right)^{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2x}}} \right]^{\frac{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{x}+\mathrm{1}}} \\ $$$$=\mathrm{e}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)::\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+\mathrm{2}^{\mathrm{x}} \mathrm{x}}{\mathrm{1}+\mathrm{3}^{\mathrm{x}} \mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \\ $$$$=\mathrm{e}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{2}^{\mathrm{x}} \mathrm{x}\right)−\mathrm{ln}\left(\mathrm{1}+\mathrm{3}^{\mathrm{x}} \mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\mathrm{e}\underset{\mathrm{x}\rightarrow\mathrm{0},\xi\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{\xi\mathrm{x}^{\mathrm{2}} }\left[\left(\mathrm{1}+\mathrm{2}^{\mathrm{x}} \mathrm{x}\right)−\left(\mathrm{1}+\mathrm{3}^{\mathrm{x}} \mathrm{x}\right)\right] \\ $$$$=\mathrm{e}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} }{\mathrm{x}} \\ $$$$=\mathrm{e}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2}^{\mathrm{x}} \mathrm{ln2}−\mathrm{3}^{\mathrm{x}} \mathrm{ln3}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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