What I'm actually trying to prove is that for every $x$ in the domain of $f$, where $f(x)=\sqrt{9-\sqrt{9-x^2}}$, the limit $\lim_{x\to a} f(x)$ exists. Now this is my approach and I would like to verify if it's correct.
$\big|\sqrt{9-\sqrt{9-x^2}}-\sqrt{9-\sqrt{9-a^2}}\big|=\bigg|\frac{\sqrt{9-a^2}-\sqrt{9-x^2}}{\sqrt{9-\sqrt{9-x^2}}+\sqrt{9-\sqrt{9-a^2}}}\bigg|\le\big|\frac{\sqrt{9-a^2}-\sqrt{9-x^2}}{\sqrt{9-\sqrt{9-a^2}}}\big|=|\frac{x^2-a^2}{\big(\sqrt{9-\sqrt{9-a^2}}\big)\big(\sqrt{9-a^2}+\sqrt{9-x^2}\big)}|$
Since $\sqrt{9-\sqrt{9-a^2}}\ge\sqrt{6}>1$,then
$|\frac{x^2-a^2}{\big(\sqrt{9-\sqrt{9-a^2}}\big)\big(\sqrt{9-a^2}+\sqrt{9-x^2}\big)}|\le|\frac{x^2-a^2}{\sqrt{9-a^2}+\sqrt{9-x^2}}|\le|\frac{x^2-a^2}{\sqrt{9-a^2}}|$
Now $x, a \in [-3, 3]$ which means $|x^2-a^2|=|x-a||x+a|\le6|x-a|$, therefore if $\epsilon >0$, we can define $\delta=\epsilon \cdot \frac{\sqrt{9-a^2}}{6}$ and thus, if $|x-a|\le \delta$: $$\big|\sqrt{9-\sqrt{9-x^2}}-\sqrt{9-\sqrt{9-a^2}}\big|\le \epsilon$$
Which proves that $\lim_{x\to a} f(x)=f(a)$
I would also like to see if there's a simpler more elegant way to quickly prove that this limit exists using only the epsilon-delta definition or maybe the squeeze theorem? Thank you!