# Epsilon-delta proof of $\lim \limits_{x \to 0} \sqrt{x^2+9} = 3$

I'm trying to prove that $\lim \limits_{x \to 0} \sqrt{x^2+9} = 3$, given $\epsilon = 0.1$

Here's what I tried:

$\lvert f(x) - 3 \rvert < \epsilon \implies \sqrt{x^2 + 9} - 3 < 0.1 > \implies \frac {x^2}{\sqrt{x^2 + 9} + 3} < 0.1$

I'm not able to make any further progress from here.

Let $$\delta = 0.1$$
If $$|x-0|<\delta$$
$$\lvert f(x) - 3 \rvert = \frac {x^2}{\sqrt{x^2 + 9} + 3} < \frac {x^2}{\sqrt{x^2 } } =|x|<0.1$$
Now the issue is to choose $\delta$ small enough that if $|x|<\delta$ then you have the desired inequality. Hint: after your simplification, the denominator is now bounded below, so it suffices to work with $Cx^2$ instead.
It will be enough to have $x^2<0.1$ and $\sqrt{x^2+9}+3\geqslant1$. But, in fact, $\sqrt{x^2+9}+3\geqslant3$. So, take $\delta=\sqrt{0.1}$.