To show that the function is continuous at $x=2$

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

1. $f(2)=15$
2. $f(x)-f(2)=3x^2+2x-16 \Rightarrow f(x)-f(2)=(x-7/3)(x+3)$

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

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

5. $7/3-\varepsilon < x < 7/3+\varepsilon$ 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.

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$3x^2+2x-16=3(x^2+\frac23x-\frac{16}3)=3(x-2)(x+\frac83)$. (Note that 2 is a root, so you should have the factor $(x-2)$ there.) –  Martin Sleziak Oct 7 '11 at 11:44
There are a few issues, including assuming the conclusion (about epsilon - you should assume |x - 2| < delta). Also, you didn't factor f(x) - 15 correctly –  The Chaz 2.0 Oct 7 '11 at 11:47
Observe that if $\varepsilon < 1/3$, $(7/3 - \varepsilon, 7/3 + \varepsilon)$ doesn't contain 2. This should be a hint that something has gone wrong. –  Hans Parshall Oct 7 '11 at 11:52
If you factor $f(x) - f(2)$ and don't get $x-2$ as one of the factors, that should be a red flag; a reason for suspicion. With a polynomial function, if you plug in $2$ and get $0$, then $x-2$ is one of the factors. –  Michael Hardy Oct 8 '11 at 3:11

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.

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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 $\delta = \frac12\delta_1$.

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Some places define continuity in the sense of Cauchy (i.e. epsilon-delta) and show it is equivalent to Heine's definition (i.e. limits). Now you need to remember two things: (1) this is a definition, not the definition. In fact the epsilon-delta one is more "correct" because it corresponds to open neighbourhoods, which is the topological definition; (2) It might not have been the definition taught wherever the OP studies, which makes sense as he attempted to approach the problem using the Cauchy definition. –  Asaf Karagila Oct 7 '11 at 11:52
@AsafKaragila: thanks, with 'the' I didn't want to point the uniqueness of the definition - I just make mistakes with articles. I aware of both definition but my solution is based on Cauchy one, I though it's clear. –  Ilya Oct 7 '11 at 11:57
@AsafKaragila: I edited my answer following your comment, thank again. –  Ilya Oct 7 '11 at 12:14