To show that the function is continuous at $x=2$ Show that $f(x)=3x^2+2x-1$ is continuous at $x=2$.


*

*$f(2)=15$

*$f(x)-f(2)=3x^2+2x-16 \Rightarrow f(x)-f(2)=(x-7/3)(x+3)$

*Let $|f(x)-f(2)| < \varepsilon\Rightarrow |(x-7/3)(x+3)| < \varepsilon$

*$|x-7/3|<\varepsilon$ or $|x+3|<\varepsilon$

*$7/3-\varepsilon < x < 7/3+\varepsilon$ thus there exists an interval $(7/3-\varepsilon, 7/3+\varepsilon)$ around 2 such that all the values lying in this interval satisfy the assumption made in 3. and $\delta=\varepsilon \Rightarrow$ the function is continuous at $x=2$.
My question: Is the above method correct? Anytime, if we have to prove that a function is continuous at $x=a$, we assume 3. and then find an interval around $x=a$ that satisfies the assumption made. 
 A: Remember the definition of continuity - you're trying to show that for any $\varepsilon > 0$ (think of this as very small) you can find some $\delta > 0$ (also very small) so that $|x - 2| < \delta$ implies directly that $|f(x) - f(2)| < \varepsilon$.
You're right to look at $|f(x) - f(2)| = |3x^2 + 2x - 16|$.  As Martin Sleziak notes in his comment, this factors as $|3(x-2)(x + 8/3)| = 3|x - 2||x + 8/3|$.  Observe that we now have a factor of $|x - 2|$ - this is great news!  The only choice we actually get to make in this whole business is how small we're choosing $|x - 2|$.  In general, if you're trying to show continuity at $a$, you should be trying to find a factor of $|x - a|$ somewhere, since you have control over this term.
Back to our problem, how do we make $3|x - 2||x + 8/3| < \varepsilon$?  We can make $|x - 2|$ as small as we like, so worry about the $|x + 8/3|$ term first.  Think of absolute value as distance: if $x$ is very close to 2, then $|x + 8/3|$ is very close to $2 + 8/3 = 14/3$.  If you choose $|x - 2| < 1$, then $x < 3$, so $|x + 8/3| < 17/3$.  So we have $3|x - 2||x + 8/3| < 3|x - 2|(17/3) = 17|x - 2|$.  If $|x - 2| < \varepsilon / 17$, then we have $17|x - 2| < 17(\varepsilon/17) = \varepsilon$ as desired.
Review the choices we made - we need $|x - 2| < 1$ and $|x - 2| < 17 / \varepsilon$.  So if you set $\delta = \min(1, 17/\varepsilon)$, then $|x - 2| < \delta$ really does imply $|f(x) - f(2)| < \varepsilon$ by the argument above.
A: Wrong from the step 4: it does not follow from 3. E.g.
$$
0.1\cdot0.2 = 0.02<0.05
$$ 
but it does not imply that $0.2<0.05$ or $0.1<0.05$.
For the continuity you can just consider 
$$
|f(2+\delta)-f(2)| = |3(2+\delta)^2+4+2\delta - 1-15| = |3\delta^2+14\delta|
$$
where the last term goes to zero with $\delta\to 0$, i.e.
$$
\lim\limits_{\delta\to 0}|f(2+\delta)-f(2)| = 0
$$
and hence
$$
\lim\limits_{\delta\to 0}f(2+\delta) = f(2)
$$
which is the definition of the continuity at $x=2$. 
Let us also say it in $\varepsilon$-$\delta$ terms. For any $\varepsilon>0$ we should provide $\delta>0$ s.t. $|f(x)-f(2)|<\varepsilon$ if $|x-2|<\delta$. Let us make denote such $x = 2+h$ for $|h|<\delta$, then
$$
|f(2+h)-f(x)| = |3h^2+14h|\leq3h^2+14|h|\leq 3\delta^2+14\delta
$$
so if we put $\delta$ such that $3\delta^2+14\delta<\varepsilon$ we prove continuity. To solve such an equation we find positive root: $\delta_1 = \frac13(-7+\sqrt{49+\varepsilon})$ and then you take $\delta = \frac12\delta_1$.
