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Forgive me if this has been asked before, but I searched and could not find an answer.

I am trying to show that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist. If the limit did exist, and was equal to $l$, then for every $\varepsilon > 0$ there would exist some $\delta > 0$ such that for all $x$,

$$0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| < \varepsilon$$

If the limit doesn't exist, then for every number $l$, there exists some $\varepsilon > 0$ such that for all $\delta > 0$,

$$0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| \geq \varepsilon$$

So, in my efforts to find some unattainable $\varepsilon$, I have found that

\begin{align} \varepsilon &\leq \left| \frac{1}{x} - l \right| \\\ &= \left| \frac{1}{x} - \frac{lx}{x} \right| \\\ &= \left| \frac{1-lx}{x} \right| \\\ &= \frac{|1-lx|}{|x|} \\\ &\implies \varepsilon |x| \leq |1 - lx| = |1 + (-lx)| \leq |1| + |-lx| \\\ &\implies 1 \geq \varepsilon |x| - |l| |x| = (\varepsilon - |l|) |x| \\\ &\implies |x| \leq \frac{1}{\varepsilon - |l|} \end{align}

So I have found that if $|x| < \frac{1}{\varepsilon - |l|}$, then $\left| \frac{1}{x} - l \right| \geq \varepsilon$, for any $\varepsilon$. But, I need this to be true for any choice of $\delta$, for some specific $\varepsilon$.

Intuitively, I can think "as $\varepsilon$ gets closer to $|l|$, then the quotient becomes larger, meaning that $|x| < \frac{1}{\varepsilon - |l|}$ is satisfied for more choices of $\delta$", but I don't know how to formalize this. Any ideas for a next step? Am I completely off base with my reasoning?

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  • $\begingroup$ How do you know that $\epsilon-|l|$ is positive (which is necessary for your last implication to hold)? $\endgroup$ – Cameron Buie Nov 27 '13 at 16:24
  • $\begingroup$ @Cameron, good point, I had missed that! $\endgroup$ – Michael Deom Nov 27 '13 at 16:34
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You're on exactly the right track. To finish up your proof, do this instead: say that for every number $L$, you'll show that for $\epsilon = 1$, you can show that for any $\delta$, there's a number $x$ with $|x - 0| < \delta$, but $|f(x) - L| > \epsilon$.

(I'll do the remainder for the case where the putatative limit is a POSITIVE number,)

Here's how: no matter what $\delta$ is, choose $x = \min(\delta, 1/(L+2))$. That makes $x$ positive.

Now $|x - 0| < \delta$ means that $x < 1/(L+2)$, so $1/x > L+2$. So $|f(x) - L|$ is at least $L+2 - L = 2 > \epsilon$.

The small insight in this proof is that you don't need to handle every possible epsilon, and that in this case, you can reasonably easily choose one particular epsilon that will suffice for the remainder of the proof, thus cleaning things up a bit.

(To handle the notion that the putative limit $L$ might be less than zero, you can pick $x$ as before, but use $\min(\delta, 1/(|L|+2)$. The gap between $1/x$ and $L$ will now be much larger -- like $2|L| + 2$ -- but that's fine.)

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  • $\begingroup$ Thank you very much! This is exactly what I was missing. I had the misconception that $|f(x) - l| \geq \varepsilon$ had to be true for every $x$ and $\delta$, when in fact it has to be true for some specific $x$ for every $\delta$. $\endgroup$ – Michael Deom Nov 27 '13 at 16:37
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For the limit of a function $f$ to exist at a point $x_0$, we need the equality of the right hand limit and the left hand limit

$$ \lim_{x\to x_0^-} f(x) = \lim_{x\to x_0^+} f(x)= \lim_{x\to x_0} f(x). $$

Check also $ \liminf $ and $ \limsup $.

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Simpler than you think, it needs no working.

Limits must be unique from both sides (where coming in from both sides makes sense).

It is easy to show that the limit coming in from + is different to the limit coming in from -, as limits must be unique (do you want a proof of this?) this limit does not exist.

If this is an assignment draw a picture and write "clearly the limit from the right and left differ" and move on.

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  • $\begingroup$ That's a completely solid proof for one of two classes of people: those who don't know much, and those who know a lot. Those who know just enough to be skeptical will say "Hey, how do you PROVE that the left limit is <something>?" and then you're back to an epsilon-delta argument. $\endgroup$ – John Hughes Nov 1 '16 at 18:56
  • $\begingroup$ @JohnHughes are the people who understand logical implication not represented in your view? $\endgroup$ – Alec Teal Nov 1 '16 at 19:15
  • $\begingroup$ Sure. They're in the "know a lot" group. THEY can translate your ambiguous "limits must be unique from both sides" into the correct statement "the limits from both sides must both exist and be equal", and can prove the theorem that if both one-sided limits exist and are equal, then so does the two-sided limit; the only easy proof I know of this requires an epsilon-delta argument...one that's trickier than the direct argument that OP asked for. (After all, OP didn't say "How should I do this?" OP said "Any ideas for a next step.") It's polite to at least answer the asked question. $\endgroup$ – John Hughes Nov 1 '16 at 19:25
  • $\begingroup$ @JohnHughes it is elementary (by that I mean first year, first term) to know "has limit =>has left and right limits that agree" and also of the contrapositive. It should also be intuitive to say that: "if the limit were to change depending on which way you come at is there obviously can be no limit", lastly, you ought to perhaps know about sequential continuity, construct two sequences, one from each side, and you can do basically the same thing to show there's no limit. It would be wise to learn that short answers do not necessarily involve skipping work, and the nature of logical implication $\endgroup$ – Alec Teal Nov 1 '16 at 19:39
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    $\begingroup$ I don't think that OP has any doubt about the truth of the statement. But OP did want an epsilon-delta proof, and I think it's reasonable to give one. As for "knowing sequential continuity", I think that's rather more a topic for a first analysis course than a first calculus course. Perhaps your students are far more sophisticated than the ones I've taught here at Brown University, however. $\endgroup$ – John Hughes Nov 1 '16 at 21:52
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All we need to show is that if we approach the limit from above and below you get a different answer.

$$\lim_{x \to 0^+} \frac{1}{x} = \infty $$

As for small positive $x$ then $\frac{1}{x}$ will be positive and getting bigger as $x$ gets closer to zero.

$$\lim_{x \to 0^-} \frac{1}{x} = -\infty $$ As for small negative $x$ then $\frac{1}{x}$ will be negative and getting more negative as $x$ gets closer to zero.

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  • $\begingroup$ Much more efficient, thank you! But I do want to get more practice proving limit propositions using the $\varepsilon$-$\delta$ definition. $\endgroup$ – Michael Deom Nov 27 '13 at 16:35

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