I was asked to prove that: $\lim_{x\to 2} (x^2-3x)=-2$ using the $\epsilon, \delta$ deifnition

I started to try and solve the epsilon inequality in this manner:

for every $\epsilon>0$ there exist $\delta >0$ such that if:

$0<|x-2|<\delta$ then $|(x^2-3x)-(-2)|<\epsilon$

Then I started to manipulate my epsilon inequality in order to get a delta in terms of epsilon by doing the following:


Now, we must see that

$$|x-1|=|x-2+1| ≤ |x-2|+1$$

And thus, that: $$|x-2||x-2+1|≤ (|x-2|+1)|x-2|$$

After this step I get stuck. I do not know how to proceed or how to use the equations I have in order to get delta in terms of epsilon.

Thanks in advance for the help.


You're almost there. If $\delta \leq 1$, and $|x-2|<\delta$, then you know $$ |x-1| \leq |x-2| +1 \leq 2 $$ and also $$ |x-1||x-2| < 2 \delta $$ In order for the right-hand side to be $\leq\epsilon$, you need to make sure that $\delta \leq \frac{\epsilon}{2}$. In order to guarantee both $\delta \leq 1$ and $\delta \leq \frac{\epsilon}{2}$, choose $\delta = \min\left\{1,\frac{\epsilon}{2}\right\}$.

  • $\begingroup$ I might be missing something but, I do not understand... why would delta be smaller or equal to one?... $\endgroup$ – Sebastián Acosta Aug 30 '18 at 16:17
  • $\begingroup$ @SebastiánAcosta: Remember we can choose $\delta$ as small as it takes to guarantee that $|f(x) - L| <\epsilon$ whenever $|x-a|<\delta$. In one part of the proof, it turns out that we need a specific (constant) upper bound for $\delta$. The number $1$ is as good as any. Later on, we see we need an upper bound for $\delta$ that is a multiple of $\epsilon$. In order for both to be satisfied, we set $\delta$ to the minimum. $\endgroup$ – Matthew Leingang Aug 30 '18 at 18:56

Since your $x$ is approaching $2$, you may assume $|x-2|< 1/2$ by choosing a positive $\delta <1/2$ in that case you have $3/2<x<5/2$ and as a result $1/2<x-1<3/2$ so $|x-1|<3/2$

Now if $|x-2|<\delta$ then $|(x-2)(x-1)| < (3/2)\delta$

Thus if your $\delta = \min \{1/2, 2\epsilon/3\}$, then $$ |x-2|<\delta \implies | (x-2)(x-1)| <\epsilon$$


You can continue as follows: $$|x-2||x-2+1|≤ (|x-2|+1)|x-2|<(\delta+1)\delta=\epsilon \Rightarrow \\ \delta^2+\delta-\epsilon=0 \Rightarrow \delta =\frac{-1+\sqrt{1+4\epsilon}}{2}.$$ Hence, for the given $\epsilon>0$, you can choose $\delta=\frac{-1+\sqrt{1+4\epsilon}}{2}>0$, so that when $0<|x-2|<\delta$, $|(x^2-3x)-(-2)|<\epsilon$.


As an alternative we can find the optimal value for $\delta$ as follow

$$|x^2-3x+2|<\epsilon \iff -\epsilon<x^2-3x+2<\epsilon$$

that is

  • $x^2-3x+2-\epsilon<0 \implies x=\frac{3\pm \sqrt{1+4\epsilon}}{2} \implies \frac{3- \sqrt{1+4\epsilon}}{2}<x<\frac{3+ \sqrt{1+4\epsilon}}{2}$
  • $x^2-3x+2+\epsilon>0 \implies x=\frac{3\pm \sqrt{1-4\epsilon}}{2} \implies x<\frac{3- \sqrt{1-4\epsilon}}{2} \land x>\frac{3+ \sqrt{1-4\epsilon}}{2}$

which corresponds to

$$|x-2|<\frac{1- \sqrt{1-4\epsilon}}{2}=\delta_{opt}$$

  • 1
    $\begingroup$ As an instructor, I don't think there's much use for an “optimal” $\delta$. Limit proofs with quadratics teach the technique of approximating in absolute value inequalities, which is much more important as one progresses in analysis. Solving the inequalities explicitly defeats that purpose. $\endgroup$ – Matthew Leingang Aug 30 '18 at 19:01
  • $\begingroup$ @MatthewLeingang I agree with you, we don't need to find that optimal value but I think that shown its derivation can be useful in order to make clear that it is convenient use some other trick to show that the limit holds. $\endgroup$ – user Aug 30 '18 at 21:28
  • $\begingroup$ Then there's also the question of how we know that $\sqrt{1-4\epsilon}$ exists at all. This depends on the intermediate value theorem and the continuity of $x\mapsto x^2$. Showing that function is continuous requires...an $\epsilon/\delta$ proof. $\endgroup$ – Matthew Leingang Aug 31 '18 at 14:54
  • $\begingroup$ @MatthewLeingang While I agree with your first observation I can't agree with this last comment. The square roots is used here to solve a quadratic inequality we don't need any deeper concept to use that. It is a simple algebraic derivation. $\endgroup$ – user Aug 31 '18 at 18:34

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