Proving $\cosh(\sqrt{1+x^2})$ is not uniformly continuous in $\mathbb{R}$

Question

Given $f(x) = \cosh(\sqrt{1+x^2})$

I am trying to show that $f(x)$ is not uniformly continuous. Specifically:

$\exists\varepsilon\ \forall\delta\ \exists x,y \in \mathbb{R}: \ |x-y| < \delta \wedge |f(x) - f(y)| \geq \varepsilon $

So far this is my thought process:

Since $\cosh(\sqrt{1+x^2}) = \frac{\exp(\sqrt{1+x^2})+\exp(-\sqrt{1+x^2})}{2}$, $f(x)$ approaches $\frac{\exp(x)}{2}$ for large enough $x$.

Intuitively I would say that since the derivative of the exponential function is not bounded and increases monotonically, no matter how small we choose $\delta$, for large enough $x$, $f(x+\delta)-f(x)$ is going to surpass any chosen $\varepsilon$.

I am trying to use mean value theorem to calculate the magnitude of $|f(x)-f(y)|$ using $f'$:

$\exists \xi \in ]x,y[:\\f(x) - f(y) = f'(\xi)\cdot(x-y)$

but by doing this, I am introducing an additional variable to the proof.

Answer 1

You have\begin{align}\lim_{x\to\infty}f'(x)&=\lim_{x\to\infty}\frac{x\sinh\left(\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}\\&=\infty.\end{align}So, if $\delta>0$, take $x>0$ such that $y\geqslant x\implies f'(x)>\frac1\delta$ and let $y=x+\frac\delta2$. Then $|x-y|=\frac\delta2<\delta$ and\begin{align}\bigl|f(x)-f(y)\bigr|&=\left|\frac{f(x)-f(y)}{x-y}\right||x-y|\\&=\bigl|f'(c)\bigr|\frac\delta2\text{ (for some $c\in(x,y)$)}\\&>\frac12,\end{align}since $f'(c)>\frac1\delta$.

Answer 2

hint

$$\cosh(\sqrt{1+x^2})=\frac 12(e^{\sqrt{1+x^2}}+e^{-\sqrt{1+x^2}})$$

$$=g(x)+h(x)$$

$$\lim_{|x|\to+\infty}h(x)=0\implies $$ $ h$ is uniformly continuous at $\Bbb R$.

So, you just need to prive that $ g $ is Not uniformly continuous .