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It is easy to prove the limit exists, all we have to show is there exists a relationship between $\delta$ and $\epsilon$. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed.

Please show through an example (you may show that $\lim_{x\rightarrow0} \frac{1}{x}$ doesn't exist)

If possible please use the explanation scheme that is used by this answer https://math.stackexchange.com/a/66552/335742

Michael Hardy
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mathnoob123
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  • Hint: negate the definition. For some $\epsilon > 0 \ \ldots$ – Sean Roberson Jul 05 '17 at 20:40
  • @SeanRoberson Are you trying to imply that I should use a proof by contradiction? If yes then maybe can you provide some example? – mathnoob123 Jul 05 '17 at 20:47
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    $$ \begin{align} & \lim_{x\to a} f(x) = L: \ & \text{For every } \varepsilon>0 \text{ there exists } \delta>0 \ & \text{ such that for every } x, \text{ if } 0<|x-a|<\delta \text{ then } |f(x) - L| < \varepsilon. \ \ & \lim_{x\to a} f(x) \ne L: \ & \text{There exists } \varepsilon>0 \text{ such that for every } \delta>0, \ & \text{ there exists } x \text{ such that } 0<|x-a|<\delta \text{ and } |f(x) - L| \ge\varepsilon. \ {} \end{align} $$ Put "For every $L$" in front of that last thing and then it says no limit exists. – Michael Hardy Jul 05 '17 at 20:49
  • Can you please provide an example by solving the limit I provided (or if you fear it's a homework then one of your own examples)? – mathnoob123 Jul 05 '17 at 20:51

2 Answers2

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Suppose $L=\lim_{x\to0}\frac{1}{x}$ is finite.

If $L>0$, then there exists $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}-L|<L$, that is, $$ 0<\frac{1}{x}<2L $$ This is a contradiction, just take $-\delta<x<0$.

Similarly if $L<0$.

Thus we can only have $L=0$. Then there should exist $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}|<1$, an obvious contradiction.

It can be neither $\lim_{x\to0}\frac{1}{x}=\infty$ nor $\lim_{x\to0}\frac{1}{x}=-\infty$, because $\frac{1}{x}$ assumes positive and negative values in every punctured neighborhood of $0$.

Can this be generalized? Not really. For instance, in order to show the non existence of $\lim_{x\to0}\sin\frac{1}{x}$ the easiest way is to show that the limit should be in the interval $[-1,1]$, but that $\sin\frac{1}{x}$ assumes every value in $[-1,1]$ in each punctured neighborhood of $0$, so it is far from every possible limit. Alternatively, there exist sequences $(a_n)$ and $(b_n)$ convergin to $0$ such that $\lim_{n\to\infty}\sin\frac{1}{a_n}=0$ and $\lim_{n\to\infty}\sin\frac{1}{b_n}=1$.

As another example, $\lim_{x\to0}e^{1/x}$ doesn't exist because the one sided limits are different: from the left it is $0$, from the right it is $\infty$.

egreg
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If $\displaystyle\lim_{x\to0} \frac 1 x = L$ then what happens if $\varepsilon=1$? There would exist $\delta>0$ such that if $0<|x|<\delta$ then $\displaystyle \left| \frac 1 x - L \right| <\varepsilon.$ But if $x$ is positive, you can make $1/x$ bigger than $L+\varepsilon$ by making $x$ less than $1/(L+\varepsilon).$ And a similar thing works if $x$ is negative (where you have $1/x\to-\infty$).

Michael Hardy
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  • Oh so the basic thing is an infinitesimal value namely $m$ will always be smaller than $\delta$ and $1/m>L+\epsilon$. Two questions. 1. Is the procedure always the same for any proof of the nonexistence of limit? 2. How to write this proof formally? 3. For limit definition (for proving a limit) is it okay to show $a\delta \leq \epsilon$ or is it always $a\delta <\epsilon$ where $a$ are the other terms that might be present? (I am aware there can be other terms, just concerned about the inequality signs?) – mathnoob123 Jul 05 '17 at 21:00
  • You need to know something about the nature of the function in order to write the proof, but what I wrote in a comment under the question is true generally. – Michael Hardy Jul 05 '17 at 21:44