Question Number 161723 by LEKOUMA last updated on 21/Dec/21
![Calculate lim_(x→0) (((1+x.2^x )/(1+x.3^x )))^(1/x^2 ) lim_(x→0) [2e^(x/(x+1)) −1]^((x^2 +1)/x) lim_(x→a) ((x^x −a^a )/(x−a))](https://www.tinkutara.com/question/Q161723.png)
$${Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{x}.\mathrm{2}^{{x}} }{\mathrm{1}+{x}.\mathrm{3}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{2}{e}^{\frac{{x}}{{x}+\mathrm{1}}} −\mathrm{1}\right]^{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{x}} −{a}^{{a}} }{{x}−{a}} \\ $$
Answered by cortano last updated on 21/Dec/21
$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+{x}.\mathrm{2}^{{x}} }{\mathrm{1}+{x}.\mathrm{3}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}^{{x}} −\mathrm{3}^{{x}} }{{x}}\right)} ={e}^{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}} \:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by cortano last updated on 21/Dec/21
$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2}{e}^{\frac{{x}}{{x}+\mathrm{1}}} −\mathrm{1}\right)^{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}} ={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left({e}^{\frac{{x}}{{x}+\mathrm{1}}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}}} \\ $$$$\:\:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}2}\left(\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }.{e}^{\frac{{x}}{{x}+\mathrm{1}}} \right)} \:=\:{e}^{\mathrm{2}} \\ $$
Commented by LEKOUMA last updated on 21/Dec/21
$${Many}\:{thanks} \\ $$
Answered by Ar Brandon last updated on 21/Dec/21
$$\mathcal{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{x}\centerdot\mathrm{2}^{{x}} }{\mathrm{1}+{x}.\mathrm{3}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \Rightarrow\mathrm{ln}\mathcal{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\mathrm{ln}\left(\frac{\mathrm{1}+{x}.\mathrm{2}^{{x}} }{\mathrm{1}+{x}.\mathrm{3}^{{x}} }\right) \\ $$$$\mathrm{ln}\mathcal{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left({x}.\mathrm{2}^{{x}} −{x}.\mathrm{3}^{{x}} \right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} −\mathrm{3}^{{x}} }{{x}}=\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\mathrm{ln}\mathcal{A}=\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\Rightarrow\begin{array}{|c|}{\mathcal{A}=\frac{\mathrm{2}}{\mathrm{3}}}\\\hline\end{array} \\ $$
Answered by Ar Brandon last updated on 21/Dec/21
$$\mathcal{B}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{x}} −{a}^{{a}} }{{x}−{a}}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}{x}^{{x}} \left(\mathrm{1}+\mathrm{ln}{x}\right)={a}^{{a}} \left(\mathrm{1}+\mathrm{ln}{a}\right) \\ $$
Commented by LEKOUMA last updated on 21/Dec/21
$${Many}\:{thanks} \\ $$