Evaluating $\lim_{x\rightarrow\infty}\left(\frac{2x-1}{3x+2}\right)^x$.

Question

I have been trying to solve this limit but i think it doesnt get me anywhere.

I tried with ln(y) but nothing.

I tried to transform it to inf/inf but no result .

Can anyone please help me find it out?

$$\lim_{x\rightarrow\infty}\left(\frac{2x-1}{3x+2}\right)^x$$

Answer 1

Following your idea we have

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

indeed

$$\log \left(\frac{x-1}{3x+2}\right)\to \log \left(\frac13\right) <0 \implies x\log \left(\frac{x-1}{3x+2}\right) \to -\infty$$

or simply

$$\frac{x-1}{3x+2}\sim \frac13 \implies \left(\frac{x-1}{3x+2}\right)^x\sim \frac1{3^x}\to 0$$

or also

$$\left(\frac{x-1}{3x+2}\right)^x=\frac1{3^x}\left(\frac{x-1}{x+2/3}\right)^x=\frac1{3^x}\left(\frac{x+2/3-5/3}{x+2/3}\right)^x=$$$$=\frac1{3^x}\left(1-\frac{5/3}{x+2/3}\right)^x\to 0\cdot e^{-5/3}=0$$

Answer 2

Compute instead the limit of the logarithm and do the substitution $x=1/t$, so the limit becomes $$ \lim_{x\to\infty}\log\left(\left(\frac{2x-1}{3x+2}\right)^x\right)= \lim_{t\to0^+}\frac{\log(2-t)-\log(3+2t)}{t} $$ Note that the numerator is negative in a neighborhood of $0$, so the limit is $-\infty$.

Since $\lim_{u\to-\infty}e^u=0$, your given limit is $0$.

Answer 3

Here is another approach. For $x\gt\frac12$ $$ \begin{align} \left(\frac{2x-1}{3x+2}\right)^x &=\left(\frac{2-\frac1x}{3+\frac2x}\right)^x\\ &\le\left(\frac23\right)^x \end{align} $$ Now the limit as $x\to\infty$ is clearer.