How can I show that $\lim\limits_{x \to \infty} \frac{\log(x+1)}{\log{x}}=1$?

Question

Hello I have a silly question: How can I show that $\lim\limits_{x \to \infty} \dfrac{\log(x+1)}{\log{x}}=1$. Thank you.

Answer 1

You could use l'Hospital's rule:

$$ \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)} = \lim_{x\to\infty} \frac{1/(x+1)}{1/x} = \lim_{x \to \infty} \frac{1}{\frac{x+1}{x}} = \lim_{x \to \infty} \frac{1}{1 +\frac{1}{x}} = \frac{1}{1 + \lim_{x \to \infty} \frac{1}{x}} = \frac{1}{1+0} = 1 $$

Answer 2

If $f$ is differentiable note that $f(x+1)-f(x)=f'(\xi)$ for some $\xi$ between $x$ and $x+1$ by the mean value theorem. Here $f(x)=\ln x$ and hence $0<f'(\xi)=\frac1\xi<\frac1x\to 0$.