Find the limit of fraction involving logarithms

Question

I am looking for a way to prove the following limit for integer $x$s:

$$\lim_{x\to\infty}{\frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)}}=\frac{1}{2}$$

I could find the result by using a computer program but I cannot formally establish the above equality.

Answer 1

You may write, as $x \to \infty$, $$ \frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)}=\frac{\log(1+\frac2x)-\log(1+\frac1x)}{\log(1+\frac2x)}=\frac{\frac1x+O(1/x^2)}{\frac2x+O(1/x^2)}=\frac12+O(1/x). $$

Answer 2

HINT : $\log(x+a)=\log(x)+\log(1+\frac{a}{x})$

Answer 3

We will use the fundamental limit $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{1}$$ We have \begin{align} L &= \lim_{x \to \infty}\frac{\log(x + 2) - \log(x + 1)}{\log(x + 2) - \log x}\notag\\ &= \lim_{x \to \infty}\dfrac{\log\left(\dfrac{x + 2}{x + 1}\right)}{\log\left(\dfrac{x + 2}{x}\right)}\notag\\ &= \lim_{x \to \infty}\dfrac{\log\left(1 + \dfrac{1}{x + 1}\right)}{\log\left(1 + \dfrac{2}{x}\right)}\notag\\ &= \lim_{x \to \infty}\dfrac{\log\left(1 + \dfrac{1}{x + 1}\right)}{\dfrac{1}{x + 1}}\cdot\dfrac{\dfrac{1}{x + 1}}{\dfrac{2}{x}}\cdot\dfrac{\dfrac{2}{x}}{\log\left(1 + \dfrac{2}{x}\right)}\notag\\ &= \frac{1}{2}\lim_{y \to 0}\frac{\log(1 + y)}{y}\cdot\lim_{x \to \infty}\frac{x}{x + 1}\cdot\lim_{z \to 0}\frac{z}{\log(1 + z)}\text{ (putting }y = 1/(x + 1), z = 2/x)\notag\\ &= \frac{1}{2}\notag \end{align}

Answer 4

The Mean Value Theorem says, $$ \begin{align} \frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)} &=\frac12\frac{\frac{\log(x+2)-\log(x+1)}{1}}{\frac{\log(x+2)-\log(x)}{2}}\\ &=\frac12\frac{\frac1{\xi_1}}{\frac1{\xi_0}}\\ &=\frac12\frac{\xi_0}{\xi_1} \end{align} $$ where $x+1\lt\xi_1\lt x+2$ and $x\lt\xi_0\lt x+2$. Therefore, $$ \frac12\frac{x}{x+2}\le\frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)}\le\frac12\frac{x+2}{x+1} $$ The Squeeze Theorem, then says that $$ \lim_{x\to\infty}\frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)}=\frac12 $$