1
$\begingroup$

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:

$$|(x^2-3x)-(-2)|=|x^2-3x+2|=|(x-2)(x-1)|=|x-2||x-1|$$

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.

$\endgroup$
2
$\begingroup$

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 $\delta = \min\left\{1,\frac{\epsilon}{2}\right\}$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$ |x-2|<\delta \implies | (x-2)(x-1)| <\epsilon$$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
-1
$\begingroup$

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

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

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.