calculus - Proof that a function is continuous in R

I have some trouble with this problem, I'll write what I did

Problem:

$ f(x) = x^2-2x $

Prove f continuous in $\Bbb R$.

My solution:

I need to prove that:
$$\lim_{x \to x_0} f(x) = f(x_0)$$

Or more specifically,

To all $\epsilon>0$ exist $\delta>0$ so to $x\in\mathbb{R}$ (function is a polynomial),

$$|x-x_0|<\delta\mbox{ then }|f(x)-f(x_0)|<\epsilon$$

So

$|f(x)-f(x_0)|$ =

$|x^2-2x-x_0^2+2x_0|$ =

$|-2(x-x_0)+(x-x_0)(x+x_0)|<\epsilon$

Now I got stuck. Any ideas?

Thanks

Answer

The idea is to bound $|x^2 -2x - x_0^2 + 2x_0|$ with stuff, so that in the end we get this to be less than $\epsilon$. We know that $|x - x_0| < \delta$. There is a very nice inequality, which we'll use: $$|a| - |b| \leq |a - b|$$
This gives us that $|x| \leq \delta + |x_0|$. There's no harm in suppose that $\delta < 1$ (why?)

So we have $|x| < 1 + |x_0|$. Now, we also use the triangle inequality. It follows: $$\begin{align} |x^2 - 2x - x_0^2 + 2x_0| &= |(x + x_0)(x -x_0) + 2(x - x_0)| \\ &= |(x + x_0 + 2)(x - x_0)| \\ &\leq (|x| + |x_0| + 2)|x - x_0| \\ &\leq (1 + |x_0| + |x_0| + 2)|x - x_0| \\ &< (3 + 2|x_0|)~\delta\end{align}$$

So, given $\epsilon > 0$, choosing $\delta = \min \left\{ 1, \frac{\epsilon}{3 + 2|x_0|}\right\}$ will solve the problem.

Once I answered a question on which I gave the general strategy for dealing with $\epsilon - \delta$ proof for polynomials like this one. Maybe you'll find it helpful.