Is my proof of "$f(x)=x^3+3x$ is continuous" correct?

Question

We have to show that the function $f(x)=x^3+3x$ is continuous at $a = 2$. By showing this it also means that that's the function's limit.

$$f(a)=L$$

$$|x-2|<\delta$$ $$x\in(\delta-2,\delta+2)$$ $$|x^3+3x-2^3-3*2|<\epsilon$$ $$|(x-2)(x^2+2x+6)-2|<\epsilon$$ $$|(x-2)(x^2+2x+6)-2|< |(x-2)|(x^2+2x+6)+|-2|<\epsilon$$ $$|x-2|<\frac{\epsilon-|-2|}{|x^2+2x+6|}=\delta$$ So then we can remove the absolute signs and get:

$$\delta = \frac{\epsilon-2}{|x^2+2x+6|}$$

End of proof. Is my proof correct? How would you make it better ? Is there one more step ? Do I need to show for $x<2$ and $x>2$ or is that already said in my proof?

Answer 1

The key inequalities are \begin{align} & |x^3 + 3x - 2^3 - 3\times 2| \\ \leq & |x^3 - 2^3| + 3|x - 2| \\ = & |x - 2||x^2 + 2x + 4| + 3|x - 2| \\ = & |x - 2|(|x^2 + 2x + 4| + 3) \\ \leq & |x - 2|(x^2 + 2|x| + 4 + 3) \\ = & (x^2 + 2|x| + 7)|x - 2| \tag{1} \end{align}

After getting this, notice the term $x^2 + 2|x| + 7$ can be bounded when $x$ is sufficiently close to $2$: say when $|x - 2| < 1$, implying $|x| < 3$. Now given $\varepsilon > 0$, if we take $\delta = \min\left(1, \frac{\varepsilon}{22}\right) > 0$, then as long as $|x - 2| < \delta$, $(1)$ can be continued as: \begin{align} |x^3 + 3x - 2^3 - 3\times 2| \leq (9 + 6 + 7)|x - 2| < 22 \times \frac{1}{22}\varepsilon = \varepsilon. \end{align}

This completes proof.

To summarize, your observation about $\delta$'s generic form $\min(a, b)$ is correct: here $a$ usually is a constant so that terms like $|x^2 + 2x + 4| + 3$ can be bounded by a fixed number, while $b$ usually depends on $\varepsilon$, which is in turn determined by the linear difference term $|x - x_0|$.

Answer 2

“Now do everything backwards”

One of the tricky aspects of an $\epsilon-\delta$ proof is that the work you do is pretty-much in reverse order to what you have to present as the proof.

Namely, you’ve done the hard work to calculate a $\delta$. Now put it all aside.

The proof starts with “Let $\epsilon > 0$. Let $\delta = $ (your calculation). We show that if $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$”.

You then write down $|f(x)-f(a)|$, and then the manipulations that conclude with it being “$ < \epsilon$”.