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Question Number 161723 by LEKOUMA last updated on 21/Dec/21

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

Answered by cortano last updated on 21/Dec/21

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

Answered by cortano last updated on 21/Dec/21

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

Commented by LEKOUMA last updated on 21/Dec/21

$$Manythanks$$

Answered by Ar Brandon last updated on 21/Dec/21

$$\mathcal{A}=\underset{x\to 0}{\lim }{\left(\frac{1+x\cdot 2^x}{1+x.3^x}\right)}^{\frac{1}{x^2}}\Rightarrow \ln \mathcal{A}=\underset{x\to 0}{\lim }\frac{1}{x^2}\ln \left(\frac{1+x.2^x}{1+x.3^x}\right)$$$$\ln \mathcal{A}=\underset{x\to 0}{\lim }\frac{1}{x^2}(x.2^x-x.3^x)=\underset{x\to 0}{\lim }\frac{2^x-3^x}{x}=\ln \left(\frac{2}{3}\right)$$$$\ln \mathcal{A}=\ln \left(\frac{2}{3}\right)\Rightarrow \overline{)\begin{array}{c}\mathcal{A}=\frac{2}{3}\end{array}}$$

Answered by Ar Brandon last updated on 21/Dec/21

$$\mathcal{B}=\underset{x\to a}{\lim }\frac{x^x-a^a}{x-a}=\underset{x\to a}{\lim }{x}^x(1+\ln x)=a^a(1+\ln a)$$

Commented by LEKOUMA last updated on 21/Dec/21

$$Manythanks$$

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