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


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.


1 Answer 1


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.

  • $\begingroup$ Technically we could choose any number to bound $\delta$ by, correct? $\endgroup$ Mar 6, 2016 at 4:02
  • $\begingroup$ Yes. We pick $1$ for simplicity. If you bound $\delta$ by a constant greater than $1$, $\epsilon/{\rm constant}$ will be even smaller, though. $\endgroup$
    – Ivo Terek
    Mar 6, 2016 at 4:04
  • $\begingroup$ Slight typo. I think you mean $<$ in the second line. $\endgroup$ Mar 6, 2016 at 16:04
  • $\begingroup$ The triangle inequality is with $\leq$, not $<$... the inequality becomes strict later, when we use $|x|<1+|a|$. $\endgroup$
    – Ivo Terek
    Mar 6, 2016 at 16:07
  • $\begingroup$ Sorry I meant fourth line. $\endgroup$ Mar 6, 2016 at 16:08

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