TITLE: Prove that $\log(1 + \sqrt{1+x^2})$ is uniformly continuous?
QUESTION [5 upvotes]: Problem  Prove that  $$\log(1 + \sqrt{1+x^2})$$ is uniformly continuous.

My idea is to consider $|x - y| < \delta$, then show that
$$|\log(1 + \sqrt{1+x^2}) - \log(1 + \sqrt{1+y^2})|
= \bigg|\log\bigg(\dfrac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}}\bigg)\bigg| < \epsilon$$
But I couldn't find a choice for $x, y$ that could implies the above expression is true. Completing the square doesn't seem to help at all. Any idea?

REPLY [4 votes]: The derivative of $f(x)=\log(1 + \sqrt{1+x^2})$ is $\frac{x}{1 + x^2 + \sqrt{1 + x^2}}$, which is bounded in the whole real line since it is continuous and tends to $0$ as $x\to\pm\infty$. By the Mean Value Theorem, $f$ is Lipschitz and so uniformly continuous.