Absolute Value Inequalities for proving a limit

Question

I am trying to prove that $\lim \limits_{x \to 0}$ $f(x) = 2x^2 -5x + 2=2$. This is what my incorrect work looks like: https://prnt.sc/1rylg4l

I know delta must be positive since it is an upper bound on distance/absolute value and therefore I must have made a mistake when solving the inequality.

I know that $|x|<1$, therefore, $-1 < x < 1 = -2 < 2x < 2 = -7 < 2x-5 < -3$.
I am trying to find the correct bound on $|2x-5|$. Thanks in advance!

Answer 1

$|2x^2- 5x+ 2- 2|= |2x^2-5x|= |x||2x- 5|$ as you have. You want $|x||2x-5|< \epsilon$ for x "small enough"- i.e. close enough to 0. Notice that your |x| is already of the form "|x- 0|", it is the |2x- 5| that is the problem.

Yes, since we are taking x close to 0 we can certainly assume that -1< x< 1 so -2< 2x< 2 and -7< 2x- 5< -3. Both bounds are negative but 3< |2x- 5|< 5.

That is we can assume that |2x- 5|< 5 so that $|x||2x- 5|< 5|x|< \epsilon$.

So we can take $\delta< \frac{\epsilon}{5}$.

Answer 2

As you pointed out, $-7 < 2x-5 < -3$, meaning $|2x-5| < 7$. The bound can't be less than 7 because then you're excluding values.

Answer 3

HINT

You can also approach it as follows.

Let $0 < |x - 0| < \delta_{\varepsilon}$. Then we conclude that \begin{align*} |f(x) - 2| & = |2x^{2} - 5x|\\\\ & \leq 2|x|^{2} + 5|x|\\\\ & < 2\delta^{2}_{\varepsilon} + 5\delta_{\varepsilon} := \varepsilon \end{align*}

Can you take it from here?