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