Limit of a function raised to a power

Question

I was working with extraction of non-electrolytic solutions and was sketching a mathematical formulae to find the limit of extracting a solvent by Nernst equation when I stumbled on this limit.

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

Can somebody explain how to obtain this limit. This might be a silly question to put up here but I couldn't find the solution with my knowledge on limits.

P.S- from experimental evidence it is evident that a limit do exist.

Answer 1

Let $$\displaystyle y= \lim_{x\rightarrow \infty}\left(\frac{2x}{2x+1}\right)^x\;,$$ Now $$\displaystyle \ln(y) = \lim_{x\rightarrow \infty}x\cdot \ln\left[\frac{2x}{2x+1}\right] = \lim_{x\rightarrow \infty}x\cdot \left[\ln(2x)-\ln(2x+1)\right]$$

Now $$\displaystyle \ln(y) = \lim_{x\rightarrow \infty}\frac{\left[\ln(2x)-\ln(2x+1)\right]}{\frac{1}{x}}$$

applying $\bf{L,Hopital\; Rule}$

$$\displaystyle \ln(y) = \lim_{x\rightarrow \infty}\frac{\frac{2}{2x}-\frac{2}{2x+1}}{-\frac{1}{x^2}} = -\lim_{x\rightarrow \infty}\frac{2x^2}{2x\cdot (2x+1)} = -\frac{1}{2}$$

so we get $$\displaystyle \ln(y) = -\frac{1}{2}\Rightarrow y = e^{-\frac{1}{2}}$$

Answer 2

Let's try answering a different question first.

Try finding $$ \lim_{x \to \infty} \left(\frac{2x+1}{2x} \right)^{2x} = \lim_{x \to \infty} \left(1+\frac{1}{2x} \right)^{2x} $$

Then see if you can transform the expression within the limit into the one you desire by applying some continuous functions. Then take the limit.

Answer 3

We have

$$\left(\frac{2n}{2n+1}\right)^n = \left(\frac{1}{1+1/(2n)}\right)^n = \left(\left(1+\frac{1/2}{n}\right)^n\right)^{-1}. $$

So $$ \lim_{n\to\infty }\left(\frac{2n}{2n+1}\right)^n =\lim_{n\to\infty } \left(\left(1+\frac{1/2}{n}\right)^n\right)^{-1}\overset{(*)}{=}\exp(1/2)^{-1} =e^{-\frac{1}{2}}.$$

In $(*)$ we used the definition of the exponential function $$\exp(x):=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$$ and the continuity of the inversion $x\mapsto 1/x$.