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Using the $\epsilon-\delta$ definition of the limit, evaluate $\lim_{x \to a} f(x)$, where $f(x) = x^3+5x$.

Attempt

We can use the usual way of constructing the limit: $$\forall \epsilon, \exists \delta \quad 0 < |x-a| < \delta \quad \implies \quad |x^3+5x-(a^3+5a)| < \epsilon.$$ Then we get $|x-a||a^2+x(a+x)+5| < \epsilon$. I imaging now we are going to have to bound $\delta$ by say $\leq 1$ to get $|x-a| < 1$. Then I'm not sure what to do next.

Ivo Terek
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user19405892
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1 Answers1

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You are in the right way. We continue the attack using the triangle inequality: $$|a^2+x(a+x)+5||x-a| \leq (|a|^2+|x|(|a|+|x|)+5)|x-a|$$

If $\delta \leq 1$, then $|x-a| <1$ implies $|x| < 1+|a|$, so: $$(|a|^2+|x|(|a|+|x|)+5)|x-a| < (|a|^2+(1+|a|)(1+2|a|)+5)|x-a|$$You can simplify that coefficient, but the point is that $\delta < \min \left\{1, \frac{\epsilon}{(|a|^2+(1+|a|)(1+2|a|)+5)}\right\}$ will work. I wrote an answer about that here, it might also help.

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Ivo Terek
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  • Technically we could choose any number to bound $\delta$ by, correct? – user19405892 Mar 06 '16 at 04:02
  • Yes. We pick $1$ for simplicity. If you bound $\delta$ by a constant greater than $1$, $\epsilon/{\rm constant}$ will be even smaller, though. – Ivo Terek Mar 06 '16 at 04:04
  • Slight typo. I think you mean $<$ in the second line. – user19405892 Mar 06 '16 at 16:04
  • The triangle inequality is with $\leq$, not $<$... the inequality becomes strict later, when we use $|x|<1+|a|$. – Ivo Terek Mar 06 '16 at 16:07
  • Sorry I meant fourth line. – user19405892 Mar 06 '16 at 16:08
  • Ah yes, then you're right. Fixed, thanks. – Ivo Terek Mar 06 '16 at 16:09
  • I just realized you can apply this same principle you applied here to pretty much any other question. – user19405892 Mar 06 '16 at 16:14
  • This line of thought is very useful.. this solves at least all questions about limits of polynomials. If you take a bit of caution with how you bound $\delta$, this works for all rational functions too. And we need not limit ourselves (pun intended) to polynomials of one variable.. I talk a bit about that in the answer I linked above – Ivo Terek Mar 06 '16 at 16:22
  • How did you get $ \frac{\epsilon}{(|a|^2+(1+|a|)(1+2|a|)+5)}$? Where did the epsilon come from? – user19405892 Mar 06 '16 at 16:37
  • I chose this because I wanted that expression in the denominator to cancel with what was multiplying $|x-a|$, leaving us with only $\epsilon$ in the end, which was what we wanted. – Ivo Terek Mar 06 '16 at 16:38