Prove that $\lim_{x\to 3} (2x^2-5)=13$

Question

Let $\epsilon > 0$ be given. Notice that $$\vert 2x^2 - 5 -13\vert <\epsilon$$

is equivalent to $$\vert 2(x-3)\vert\cdot\vert(x+3)\vert<\epsilon$$

Now because the domain of the function is $\Bbb{R^1}$, we can restrict the interval $a-\delta<x< a +\delta$ to $\delta < 6$. $$\vert x-3\vert < 6$$ $$-6 < x-3 <6$$ add 6 $$0<x+3<12$$ $$\vert 2(x+3) \vert < 24$$

This implies that $$\vert 2(x+3) \vert\cdot \vert(x-3)\vert< \vert(x-3)\vert24<\epsilon$$

$$\vert x-3 \vert < \frac{\epsilon}{24}$$

Now just choose $\delta= 6, \frac{\epsilon}{24}$, whichever happens to be the smallest. Is this correct?

Answer 1

No, it isn't correct. This is not a proof of the limit. Instead, it is the scratch work one does to figure out how one proves the limit. Now that you've done the scratch work, you put it aside (it is not part of the proof itself), and write the actual proof.

By definition, $$\lim_{x\to 3} (2x^2-5)=13$$ means:

$$\forall \epsilon > 0, \exists \delta > 0,\forall x, |x - 3| < \delta \implies |(2x^2 -5) -13| < \epsilon$$

To prove the limit, you have to show this statement is true. The proof should look something like this:

Let $\epsilon > 0$, and let $\delta$ be the smaller of $6$ and $\epsilon/24$. Let $x$ be a real number. If $|x - 3| < \delta$, then [insert main argument here], so $|(2x^2 -5) -13| < \epsilon$.

(Exact wording is of course a matter of taste, but the proof must follow the same general flow). Its elements are:

• "Let $\epsilon > 0$," - we start with an arbitrary $\epsilon$ meeting the condition $> 0$. Since it is arbitrary, the remainder of the proof works no matter what $\epsilon$ is picked, thus showing that the "for all" quantifier is satisfied.
• "and let $\delta$ be the smaller of $6$ and $\epsilon/24$" - since the quantifier for $\delta$ is "there exists", we can specify a particular $\delta$. However, the value of $\delta$ chosen can only depend on $\epsilon$: it cannot depend on $x$, which has not been introduced yet. This is what your scratch work was for - to figure out what $\delta$ we should choose at this stage. It can also guide you with the main argument - but that argument is going to be in the reverse direction from your scratch work. The scratch work itself is not a part of this proof.
• "Let $x$ be a real number" - again, an arbitrary $x$ will allow us to say that the "for all" quantifier is satisfied.
• "If $|x - 3| < \delta$, then" - this introduces the condition on $x$ that allows us to reach a conclusion. Because of the way you've chosen $\delta$, you know two things: $$|x - 3| < 6\\|x - 3| < \dfrac \epsilon {24}$$
• [insert main argument here] - This is a calculation that starts with the two facts about $x$ above, and from them derives $|(2x^2 -5) -13| < \epsilon$. Note that your scratch work went in the other direction. You will have to reverse the flow. But this is not necessarily just writing it in reverse order. Some of the things you did in the scratch work may not be reversible, so you need to be careful here.
• "so $|(2x^2 -5) -13| < \epsilon$" - This is the conclusion of your main argument. It is the destination the argument is headed for. It is what you need to show.

Answer 2

Since $\lim_{x\to 3} (2x^2-5)$ is not one of the indeterminate forms, you can find the limit just by evaluating $f(x)= 2x^2-5$ at $x=3.$