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

$$Calculate$$$$1.\underset{x\to 0}{\lim }{[2e^{\frac{x}{x+1}}-1]}^{\frac{x^2+1}{x}}$$$$2.\underset{x\to 0}{\lim }{\left(\frac{1+x\times 2^x}{1+x\times 3^x}\right)}^{\frac{1}{x^2}}$$$$$$

Answered by qaz last updated on 02/Dec/21

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

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