Show that function $f(x)=\sqrt{x^2+1}$ is continuous using Cauchy's definition of continuity.

Question

Show that function $f(x)=\sqrt{x^2+1}$ is continuous at $0$ using Cauchy's definition of continuity.

I still haven't fully grasped this defition, but if I got this correct, I need to find $\delta$ that depends on any given $\epsilon$ such that if $|x-a|=|x-0|<\delta$, then $|f(x)-f(a)|=|f(x)-1|<\epsilon$. How do I connect these two inequalities?

I started with second inequality and got rid of the square root but I dont have any other ideas.

$|f(x)-1|=|\sqrt{x^2+1}-1|=|\frac{x^2+1-1}{\sqrt{x^2+1}+1}|=|\frac{x^2}{\sqrt{x^2+1}+1}|$

Answer 1

\begin{align*} \dfrac{x^{2}}{\sqrt{x^{2}+1}+1}\leq x^{2}\leq\delta^{2}<\epsilon, \end{align*} by choosing $\delta=\sqrt{\epsilon}$.

Answer 2

A little trickery:

$|\sqrt{x^2+1} -1| \dfrac{\sqrt{x^2 +1} +1}{\sqrt{x^2+1}+1} =$

$\dfrac{x^2}{\sqrt{x^2+1}+1}\le$

$\dfrac{x^2}{|x|} =|x|.$

For a given $\epsilon \gt 0$ , choose $\delta = \epsilon.$

Note: $\sqrt{x^2+1} +1 \gt \sqrt{x^2} = |x|.$