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I'm a relative novice to epsilon-delta proofs. My professor assigned this practice problem and I'm having terrible trouble understanding the answer he gave. Moreover, I can't find a good account for a general strategy for how to do these sorts of proofs. I understand the epsilon delta definition, I understand what I'm supposed to do, but I need advice on strategy.

The actual problem.

Prove that the limit

$$ \lim_{x \rightarrow 2} \frac{x^3}{x-2} $$ does not exist.

So far I'm pretty stumped; I know I need to show that there is some $\epsilon$ st. such that x being arbitrarily close to 2 does not guarantee that f(x) is within epsilon of L, but that's all I've got.

Thanks.

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  • $\begingroup$ Do you understand intuitively why this limit doesn't exist? $\endgroup$ – Henning Makholm Nov 13 '16 at 0:04
  • $\begingroup$ Yes; it's unbounded at 2. $\endgroup$ – BenL Nov 13 '16 at 0:06
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    $\begingroup$ Couple of hints in no particular order: $f(x)$ changes sign in any neighborhood of $2$. Also, $f(x)$ is unbounded in any neighborhood of $2$. $\endgroup$ – dxiv Nov 13 '16 at 0:10
  • $\begingroup$ Uniqueness of limits has not been proven in class and thus can't be used in this proof. And I know it's unbounded; I just don't have any idea how to formalize this into a proof. $\endgroup$ – BenL Nov 13 '16 at 0:13
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    $\begingroup$ Can an unbounded function stay within a (bounded) neighborhood $(L - \epsilon, L + \epsilon)$? $\endgroup$ – dxiv Nov 13 '16 at 0:15
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First, let's simplify the problem. The numerator has a greater degree than the denominator, which means we can use long division: $$\frac{x^3}{x-2} \equiv x^2+2x+4 + \frac{8}{x-2}$$ Standard properties of limits, i.e. $\lim (\mathrm f+\mathrm g) = \lim \mathrm f + \lim \mathrm g$ and $\lim (k\mathrm f) = k \lim \mathrm f$, give $$\lim_{x \to 2} \left(\frac{x^3}{x-2}\right) = 12+\lim_{x \to 2} \left(\frac{8}{x-2}\right) = 12+8\, \lim_{x \to 2} \left(\frac{1}{x-2}\right)$$

Clearly the limit of $\frac{1}{x-2}$ as $x \to 2$ is not defined and the original limit is not defined.

It's enough to show that $\frac{1}{x-2}$ is unbounded.

For any $L>0$, we can find $x >2$ for which $\frac{1}{x-2} > L$.

If $\frac{1}{x-2} > L$ then $0 < x-2 < \frac{1}{L}$, and so $2< x < 2 + \frac{1}{L}$.

How big do you want $\frac{1}{x-2}$ to be? Let's say $L = 1,000,000$. For any $2 < x< 2 + \frac{1}{L}$, i.e. $2 < x < 2.000001$, you'll have $\frac{1}{x-2} > 1,000,000$.

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a limit exists iff $\exists a$ $\forall \epsilon$ $\exists \delta$ so that $|x-2|<\delta \implies |f(x)-a|<\epsilon$

To prove a limit does not exist, you need to prove the opposite proposition, i.e.

$\forall a$ $\exists \epsilon$ $\forall \delta$ so that $\exists x,|x-2|<\delta$ and $|f(x)-a|>\epsilon$

$2+\delta/2$ and $2-\delta/2$ are good candidates : they are close enough from 2, and their image by the function are really far apart, so one of them will be far away enough from your $\epsilon$

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Recall the definition of limit as follow:

We write $ \lim\limits_{x \to 2} f(x) = a $ if for any $ \epsilon > 0 $, there exists $ \delta $, possibly depending on $ \epsilon $, such that $ |f(x) - a| < \epsilon $ for all $ x $ such that $ |x - 2| < \delta $.

Now we look at $ \frac{x^3}{x - 2} $, the closer it gets to 2 (for example, 1.99 or 2.01), the larger it is and it can grow without bound. That means for any interval that contains 2, the value of $ f $ get infinitely large, then how can we make it arbitrarily close to any particular value?

This contradiction indicates the limit cannot possibly exists!

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    $\begingroup$ You've got the definition of limit backwards ($\delta$ depends on $\epsilon$, there is no such $\delta$ for $\forall \epsilon$ unless the function is constant). $\endgroup$ – dxiv Nov 13 '16 at 1:17
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    $\begingroup$ Thank you for your comment! That is also my understanding, I updating my wording to reflect that. $\endgroup$ – Andrew Au Nov 13 '16 at 3:39

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