Question Number 140000 by bramlexs22 last updated on 03/May/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+\mathrm{x}.\mathrm{2}^{\mathrm{x}} }{\mathrm{1}+\mathrm{x}.\mathrm{3}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} =? \\ $$
Commented by MJS_new last updated on 03/May/21
$$\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\frac{\mathrm{4}}{\mathrm{9}}\:\mathrm{I}\:\mathrm{think} \\ $$
Commented by bramlexs22 last updated on 03/May/21
$$\mathrm{yes}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$
Answered by TheSupreme last updated on 03/May/21
![(((1+ln(2)x^2 )/(1+ln(3)x^2 )))^(2/x^2 ) u=(1/x^2 ) lim (((1+((ln(2))/u^2 ))/(1+((ln(3))/u^2 ))))^(2u^2 ) =lim(((u^2 +ln(2))/(u^2 +ln(3))))^(2u^2 ) =lim(1+((ln(2)−ln(3))/(ln(3)+u^2 )))^(2u^2 ) v=u^2 +ln(3) lim (1+((ln(2)−ln(3))/v))^(2v−2ln(3)) [e^(2/(ln(2)−ln(3))) ]^](https://www.tinkutara.com/question/Q140006.png)
$$\left(\frac{\mathrm{1}+{ln}\left(\mathrm{2}\right){x}^{\mathrm{2}} }{\mathrm{1}+{ln}\left(\mathrm{3}\right){x}^{\mathrm{2}} }\right)^{\frac{\mathrm{2}}{{x}^{\mathrm{2}} }} \\ $$$${u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${lim}\:\left(\frac{\mathrm{1}+\frac{{ln}\left(\mathrm{2}\right)}{{u}^{\mathrm{2}} }}{\mathrm{1}+\frac{{ln}\left(\mathrm{3}\right)}{{u}^{\mathrm{2}} }}\right)^{\mathrm{2}{u}^{\mathrm{2}} } ={lim}\left(\frac{{u}^{\mathrm{2}} +{ln}\left(\mathrm{2}\right)}{{u}^{\mathrm{2}} +{ln}\left(\mathrm{3}\right)}\right)^{\mathrm{2}{u}^{\mathrm{2}} } ={lim}\left(\mathrm{1}+\frac{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{3}\right)+{u}^{\mathrm{2}} }\right)^{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$${v}={u}^{\mathrm{2}} +{ln}\left(\mathrm{3}\right) \\ $$$${lim}\:\left(\mathrm{1}+\frac{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)}{{v}}\right)^{\mathrm{2}{v}−\mathrm{2}{ln}\left(\mathrm{3}\right)} \\ $$$$\left[{e}^{\frac{\mathrm{2}}{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)}} \right]^{} \\ $$
Answered by bobhans last updated on 03/May/21
$$\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+\mathrm{x}.\mathrm{2}^{\mathrm{x}} }{\mathrm{1}+\mathrm{x}.\mathrm{3}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}.\mathrm{2}^{\mathrm{x}} \right)−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}.\mathrm{3}^{\mathrm{x}} \right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:\mathrm{L}=\mathrm{2}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)\right)−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{3}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:\mathrm{L}=\mathrm{2}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)−\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{3}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)\right)\right.}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\mathrm{2}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{ln}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\mathrm{ln}\:\left(\frac{\mathrm{4}}{\mathrm{9}}\right)\:\Rightarrow\:\mathrm{L}\:=\:\mathrm{e}^{\mathrm{ln}\:\left(\frac{\mathrm{4}}{\mathrm{9}}\right)} \\ $$$$\mathrm{L}\:=\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$