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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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4 Answers 4

up vote 0 down vote accepted

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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We know that for this particular situation , where x->infinity and the indeterminate form is 1 ^Infinity, we cand use Euler's limit : (1+1/u(x))^u(x)=e .

Our example needs a little more polishing . Therefore , by adding 1 and substract 1 from the base function , we arrive here (1 + 1/(3x-2)/4)^(2x) . Next up , we continue by building our exponent so that it matches the Euler identity . Since u(x)=(3x-2)/4 is , we multiply and divide our old exponent ( 2x ) by u(x) . Hence , we arrive at e^(2x/u(x)) ,which is obviosuly exp(8/3).I'm terriby sorry for not using LaTex ,mate, I'm new here . Hope you'll understand my soultion . Cheers !

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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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