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Ok so I have been trying for days already to find a solution to this all around the web and in math books but to no success. The problem is to evaluate a limit of a function composed by polynomial functions and an exponential function:

$$\lim_{x \to +\infty} \left(\frac{3x+2}{3x-2}\right)^{2x}$$

I know from a software that the solution is $\exp\left(\dfrac{8}{3}\right)$, but I can't reach this.

One thing I did to try to find the limit was:

$$\lim_{x \to +\infty}\left(\frac{3x+2}{3x-2}\right)^{2x}=\lim_{x \to +\infty}\exp\left(\ln\left(\left(\frac{3x+2}{3x-2}\right)^{2x}\right)\right)\\ =\lim_{x \to +\infty}\exp\left(2x\ln\left(\frac{3x+2}{3x-2}\right)\right) =\exp{\left(\lim_{x \to +\infty}2x\times\lim_{x \to +\infty}\ln\left(\frac{3x+2}{3x-2}\right)\right)}$$

But this doesn't work because the limit to the right goes to zero while the one on the left goes to infinity. I tried other things too, but the problem only gets more complicated and a solution seems to get farther and farther away.

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3 Answers 3

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Write $$ \frac{3x+2}{3x-2}=\frac{3x-2+4}{3x-2}=1+\frac{4}{3x-2} $$ and then make the substitution $t=3x-2$, so $x=(t+2)/3$ and your limit becomes $$ \lim_{t\to\infty}\left(1+\frac{4}{t}\right)^{2\frac{t+2}{3}}= \left(\lim_{t\to\infty} \left(1+\frac{4}{t}\right)^t\cdot \lim_{t\to\infty}\left(1+\frac{4}{t}\right)^2 \right)^{2/3} $$

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Hint: $$\lim_{x \to +\infty}\exp\left(2x\ln\left(\frac{3x+2}{3x-2}\right)\right)=\exp\left(2\lim_{t \to 0}\dfrac{\ln\left(\dfrac{3+2t}{3-2t}\right)}{t}\right)$$

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To get a nicer form, use $$ \left(\frac{3x+2}{3x-2}\right)^{2x} =\left(\frac{\left(1+\frac{2}{3x}\right)^x}{\left(1-\frac{2}{3x}\right)^x}\right)^2 $$ to get the exponential limit twice in standard form.

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