Prove that $$\lim_{x\to 3}{(x^2+2x-1)\over \sqrt{x+1}} = 7.$$
I've been stuck on this for over an hour now. I've tried using the definition of a limit and reducing $|f(x)- L|$ to make it look like $(x-3)$, but I can't seem to do it. I've also tried bounding $1\over \sqrt{x+1}$, but that doesn't seem to work either.
Note that $x^2+2x-1=(x+5)(x-3)+14$. Therefore \begin{align*}\frac{(x^2+2x-1)}{\sqrt{x+1}}-7 &=\frac{(x+5)(x-3)}{\sqrt{x+1}} +\frac{7(2-\sqrt{x+1})}{\sqrt{x+1}}\\ &=\frac{(x+5)(x-3)}{\sqrt{x+1}} +\frac{7(4-(x+1))}{\sqrt{x+1}(2+\sqrt{x+1})}\\ &=\left([x+5 -\frac{7}{(2+\sqrt{x+1})}\right)\frac{(x-3)}{\sqrt{x+1}} . \end{align*} It follows that if $x\in (0,4)$ then $$\left|\frac{(x^2+2x-1)}{\sqrt{x+1}}-7\right| \leq \left(4+5 +\frac{7}{(2+1)}\right)|x-3|.$$ Can you take it from here?