Stuck with proving $\lim_{x\to 2} (x^2-3x)=-2$, using the $\epsilon, \delta$ deifnition

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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