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Question Number 140000 by bramlexs22 last updated on 03/May/21

$$\underset{x\to 0}{\lim }{\left(\frac{1+\mathrm{x}.2^{\mathrm{x}}}{1+\mathrm{x}.3^{\mathrm{x}}}\right)}^{\frac{2}{{\mathrm{x}}^2}}=?$$

Commented by MJS_new last updated on 03/May/21

$$\mathrm{the}\mathrm{answer}\mathrm{should}\mathrm{be}\frac{4}{9}\mathrm{I}\mathrm{think}$$

Commented by bramlexs22 last updated on 03/May/21

$$\mathrm{yes}\mathrm{the}\mathrm{answer}\mathrm{is}\frac{4}{9}$$

Answered by TheSupreme last updated on 03/May/21

$${\left(\frac{1+ln\left(2\right)x^2}{1+ln\left(3\right)x^2}\right)}^{\frac{2}{x^2}}$$$$u=\frac{1}{x^2}$$$$lim{\left(\frac{1+\frac{ln\left(2\right)}{u^2}}{1+\frac{ln\left(3\right)}{u^2}}\right)}^{2u^2}=lim{\left(\frac{u^2+ln\left(2\right)}{u^2+ln\left(3\right)}\right)}^{2u^2}=lim{(1+\frac{ln\left(2\right)-ln\left(3\right)}{ln\left(3\right)+u^2})}^{2u^2}$$$$v=u^2+ln\left(3\right)$$$$lim{(1+\frac{ln\left(2\right)-ln\left(3\right)}{v})}^{2v-2ln\left(3\right)}$$$${\left[e^{\frac{2}{ln\left(2\right)-ln\left(3\right)}}\right]}^{}$$

Answered by bobhans last updated on 03/May/21

$$\mathrm{L}=\underset{x\to 0}{\lim }{\left(\frac{1+\mathrm{x}.2^{\mathrm{x}}}{1+\mathrm{x}.3^{\mathrm{x}}}\right)}^{\frac{2}{{\mathrm{x}}^2}}$$$$\ln \mathrm{L}=2\underset{x\to 0}{\lim }\frac{\ln (1+\mathrm{x}.2^{\mathrm{x}})-\ln (1+\mathrm{x}.3^{\mathrm{x}})}{{\mathrm{x}}^2}$$$$\ln \mathrm{L}=2\underset{x\to 0}{\lim }\frac{\ln (1+\mathrm{x}+{\mathrm{x}}^2\ln 2+\mathrm{o}\left({\mathrm{x}}^2\right))-\ln (1+\mathrm{x}+{\mathrm{x}}^2\ln 3+\mathrm{o}\left({\mathrm{x}}^2\right))}{{\mathrm{x}}^2}$$$$\ln \mathrm{L}=2\underset{x\to 0}{\lim }\frac{(\mathrm{x}+{\mathrm{x}}^2\ln 2+\mathrm{o}\left({\mathrm{x}}^2\right)-(\mathrm{x}+{\mathrm{x}}^2\ln 3+\mathrm{o}\left({\mathrm{x}}^2\right))}{{\mathrm{x}}^2}$$$$\ln \mathrm{L}=2\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2\left(\ln \left(\frac{2}{3}\right)\right)}{{\mathrm{x}}^2}$$$$\ln \mathrm{L}=\ln \left(\frac{4}{9}\right)\Rightarrow \mathrm{L}={\mathrm{e}}^{\ln \left(\frac{4}{9}\right)}$$$$\mathrm{L}=\frac{4}{9}$$

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