Proving Limit of Rational Function with Quadratic Numerator and Radical Denominator

Question

$$Prove \quad\lim_{x \to 3} \frac{x^2+2x-1}{\sqrt{x+1}} = 7$$

$|f(x)-L|$ is $$\frac{x^2+2x-1-7\sqrt{x+1}}{\sqrt{x+1}}$$

and I am not sure how I should proceed from here.

The quadratic isn't factorable and I don't see where I would get $|x-3|$ from.

Answer 1

The function $f(x)=\frac{x^2+2x-1}{\sqrt{x+1}}$ is continuous at $x=3$, since it is an elementary function and well defined at $x=3$, so that limit is just $f(3)=7$

Answer 2

If you are allowed to assert the equality

$$\lim_{x\to3}{x^2+2x-1\over\sqrt{x+1}}=\lim_{u\to4}{(u^2-1)^2+2(u^2-1)-1\over\sqrt{(u^2-1)+1}}$$

as a step in a proof, then your best bet may be to start with that. Eliminating the square root makes life a lot easier.