square limit by using the definition

Let $f(x)=\sqrt{x^2+1}-\sqrt{x^2-1}.$ By using the definition of limit show that $$\lim_{x\to\infty}f(x)=0$$

Easy calculation shows that $\lim_{x\to\infty}f(x)=0.$ Recall the definition $$ \text{for all} \ \ \epsilon>0 \ \text{there exists}\ M\ \text{such that}\ \ |f(x)-0|< \epsilon \ \text{for all}\ x>M$$ Now, $$|\sqrt{x^2+1}-\sqrt{x^2-1}|=|\frac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}}|<|\frac{2}{\sqrt{x^2+1}}|<|\frac{2}{\sqrt{x^2}}|=\frac{2}{|x|}$$ Take $M=\frac{2}{\epsilon}$

Am I missing something?

Thank you in advance.

EDIT. The OP was incorrect as mentioned in the comment below by @J.G. I fixed

Your correcting edit obtains basically the expected argument. As @aschepler notes, one should take e.g. $M=\max\{1,\,\frac{2}{\epsilon}\}$ so that $f(x)$ exists for all $x>M$. This constraint is sufficient for the rest of the proof.