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What is the answer to this:

$$ \lim_{x\to ∞} \left({2x+3\over 2x-1}\right)^x $$

My calculator says this is $ e^2 $ but the only answer I can get to is $ 1^\infty $, which is indeterminate.

Kamil Jarosz
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user265554
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    The fact that "naive" computation of the limit leads to an indeterminate form does not prevent the limit from possibly existing. For example, $\lim_{x\to 0}\frac{\sin x}{x}$ is also indeterminate of form $\frac 00$, but the limit exists. – Hagen von Eitzen Feb 11 '16 at 15:51
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    The fact that you get an indeterminate form does not mean the limit does not exist. In fact, $$e=\lim_{x\to\infty}\left(1+\frac1x\right)^x$$ comes from an indeterminate form. –  Feb 11 '16 at 15:51
  • I know, it's just the only answer that I got to when naively trying to calculate the limit. I'd like to know why the limit is e^2. – user265554 Feb 11 '16 at 15:52
  • Indeterminate means precisely that you need to compute the limit. – orion Feb 11 '16 at 15:58

5 Answers5

6

$$\lim_{x\to+\infty}\left(\frac{2x+3}{2x-1}\right)^x=\lim_{x\to+\infty}\left(\frac{2x-1+4}{2x-1}\right)^x=\lim_{x\to+\infty}\left(1+\frac{4}{2x-1}\right)^x=$$ $$\lim_{x\to+\infty}\left(\left(1+\frac{1}{\frac{2x-1}{4}}\right)^\frac{2x-1}{4}\right)^\frac{4x}{2x-1}=e^2$$

since:

I. $$\lim_{x\to+\infty}\frac{2x-1}{4}=+\infty$$

and then $$\lim_{x\to+\infty}\left(1+\frac{1}{\frac{2x-1}{4}}\right)^\frac{2x-1}{4}=e$$

II. $$\lim_{x\to+\infty}\frac{4x}{2x-1}=\lim_{x\to+\infty}\frac{4}{2-\frac{1}{x}}=2$$

Leon
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Using the limit definition of the exponential function

$$e^z=\lim_{x\to \infty}\left(1+\frac zx\right)^x$$

we can write

$$\begin{align} \lim_{x\to \infty}\left(\frac{2x+3}{2x-1}\right)^x&=\lim_{x\to \infty}\left(\frac{1+\frac{3/2}{x}}{1+\frac{-1/2}{x}}\right)^x\\\\ &=\frac{\lim_{x\to \infty}\left(1+\frac {3/2}{x}\right)^x}{\lim_{x\to \infty}\left(1+\frac {-1/2}{x}\right)^x}\\\\\ &=\frac{e^{3/2}}{e^{-1/2}}\\\\ &=e^2 \end{align}$$

Mark Viola
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Consider $$A= \left({2x+3\over 2x-1}\right)^x$$ $$\log(A)=x\log\left({2x+3\over 2x-1}\right)=x\log\left(1+{4\over 2x-1}\right)$$ Now, remembering that, for small $y$, $\log(1+y)\sim y$ $$\log(A)\sim x \times {4\over 2x-1}={4x\over 2x-1}\sim {4x\over 2x}=2$$

Claude Leibovici
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$$\left(\frac{2x+3}{2x-1}\right)^x=\left(\frac{1+\frac{3}{2x}}{1+\frac{-1}{2x}}\right)^x=\frac{\left[\left(1+\frac{1}{\frac{2x}{3}}\right)^{\frac{2x}{3}}\right]^{\frac{3}{2}}}{\left[\left(1+\frac{1}{-2x}\right)^{-2x}\right]^{\frac{-1}{2}}}$$

Now, noting that$\left(1+\frac{1}{y}\right)^y\to e$ as $\vert y \vert \to \infty$, we can take $y=\frac{2x}{3}, -2x$ in the numerator and denominator respectively to see that this limit tends to $\frac{e^{\frac{3}{2}}}{e^{\frac{-1}{2}}}=e^2$ as required.

πr8
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$$\left(\frac{2x+3}{2x-1}\right)^x=\exp\left(x\ln\left(1+\frac{4}{2x-1}\right)\right)$$

We have $\ln\left(1+\frac{4}{2x-1}\right)=\frac{4}{2x-1}+o_{+\infty}\left(\frac{1}{x}\right)$.

Then $x\ln\left(1+\frac{4}{2x-1}\right)=\frac{4x}{2x-1}+o(1)\to 2$ as $x\to+\infty$.

Hence $\lim\limits_{x\to +\infty}\left(\frac{2x+3}{2x-1}\right)^x=e^2$

Augustin
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