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Find the limit $$ \lim_{x \to 1}\left(\frac{3x}{2+x}\right)^\frac{x}{1-x} $$

I've transformed the function by changing limit of 1 to zero and become the following: $$ \lim_{x \to 0}\left(1 + \frac{3x-3}{x-3}\right)^\frac{1 - x}{x} $$

I spent many time and tried to transform the power part $\frac{1 - x}{x}$ to $\frac{x-3}{3x-3}$ but still can't remove the $x$'s on final. How can I find this limit? Thanks a lot!

mathstudent1
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Bi Ao
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  • It must be something wrong here with your change: setting $x=0$ in the second expression results in $1+1=2$ in the parenthesis, however it was $1$ earlier. – A.Γ. Jan 03 '21 at 10:16

3 Answers3

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$$ L=\lim_{x \to a}(f(x))^{g(x)} $$ So, if $\lim_{x \to a}(f(x))=1,\lim_{x \to a}(g(x))= \infty $ $$ \ln L=\lim_{x \to a}(f(x)-1)({g(x)}) $$ In your case, $$ L=\lim_{x \to 1}(\frac{3x}{2+x})^{\frac{x}{1-x}} $$ Therefore, $$ \ln L=\lim_{x \to 1}(\frac{3x}{2+x}-1)({\frac{x}{1-x}}) $$ $$ \ln L=\lim_{x \to 1}(\frac{2x-2}{2+x})({\frac{x}{1-x}}) $$ $$ \ln L=\lim_{x \to 1}(\frac{-2x}{2+x}) $$ $$ L= e^{\frac{-2}{3}} $$

Proof of the first claim is here: 1 to the power of infinity formula

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Hint: try to match $$ \frac{3x}{2+x}=1+\frac1t\iff t=\frac{x+2}{2(x-1)}. $$ Now $$ \frac{x}{1-x}=t\cdot\frac{-2x}{x+2}. $$ What is the limit of $$ \left[\left(1+\frac1t\right)^t\right]^{-\frac{2x}{x+2}}? $$

A.Γ.
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$$A=\Bigg[\frac{3x}{2+x}\Bigg]^\frac{x}{1-x}$$ Make $x=1+y$ to work around $y=0$. So $$A=\Bigg[\frac{3y+3}{y+3}\Bigg]^{-\frac{y+1}{y}}=\Bigg[3-\frac{6}{y+3}\Bigg]^{-\frac{y+1}{y}}$$ $$\log(A)=-\frac{y+1}{y}\log\Bigg[3-\frac{6}{y+3}\Bigg]$$ Now, by Taylor $$\log\Bigg[3-\frac{6}{y+3}\Bigg]=\frac{2 y}{3}-\frac{4 y^2}{9}+O\left(y^3\right)$$ $$\log(A)=-\frac{2}{3}-\frac{2 y}{9}+O\left(y^2\right)$$ $$A=e^{\log(A)}=\frac{1}{e^{2/3}}-\frac{2 y}{9 e^{2/3}}+O\left(y^2\right)$$

Claude Leibovici
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