MHB Assume the Limit Exists: Proving an Impossibility

[Amad27]

Hello,

Prove that

$$\lim_{{x}\to{0}} \frac{1}{x}$$

Does not exist by contradiction. So the obvious step:Assume:$$\lim_{{x}\to{0}} \frac{1}{x} = M$$

$| 1/x - M| < \epsilon$ for $|x| <\delta_1$

Any ideas? PLEASE DO NOT SOLVE.

 

[Fallen Angel]

If this limit exists, then it wouldn't matter if you go to zero from positive or from negative numbers.

 

[Amad27]

If this limit exists, then it wouldn't matter if you go to zero from positive or from negative numbers.

So I suppose what you are suggesting is that we reach a contradiction using sided limits? So we get:

$$x< \delta \implies \left| 1/x - L \right| < \epsilon$$

$$-x < \delta_2 \implies \left| 1/x - L \right| < \epsilon$$

Let $\delta' = \min(\delta_1, \delta_2)$

Let $\epsilon = 1$

$$x< \delta \implies \left| 1/x - L \right| < 1$$

$$-x < \delta_2 \implies \left| 1/x - L \right| < 1$$

------------------------------------------------------------------

$$0 < \delta_1 + \delta_2 \implies 2|1/x - L| < 2$$

This is hard, since I cannot reach a contradiction.

 

[Siron]

EDIT:
In which sense do you want to prove the limit does not exist? Do you want to prove the two-sides limit does not exist? Or do you want to prove that $\lim_{x \to 0^{+}} \frac{1}{x} = +\infty$ and $\lim_{x \to 0^{-}} \frac{1}{x} = - \infty$ which also means that both limits do not exist as they're not finite.

If it's the first case then post #2 gives the answer. If not, you could prove it this way. Suppose you want to prove $\lim_{x \to 0^{+}} \frac{1}{x}$ does not exists (in the sense that it's value is not finite) then it's equivalent with proving $\forall M \in \mathbb{R}: \lim_{x \to 0^{+}} \frac{1}{x} \neq M$. This last statement is equivalent with
$$\forall M \in \mathbb{R}: \exists \varepsilon>0, \forall \delta>0, \exists x: 0<x<\delta \ \mbox{and} \left |\frac{1}{x}-M \right| \geq \varepsilon$$

To prove the above statement. Let $\epsilon = 1$ for example and try to find an $x$ such that the above statement holds.

 

[Amad27]

EDIT:
In which sense do you want to prove the limit does not exist? Do you want to prove the two-sides limit does not exist? Or do you want to prove that $\lim_{x \to 0^{+}} \frac{1}{x} = +\infty$ and $\lim_{x \to 0^{-}} \frac{1}{x} = - \infty$ which also means that both limits do not exist as they're not finite.

If it's the first case then post #2 gives the answer. If not, you could prove it this way. Suppose you want to prove $\lim_{x \to 0^{+}} \frac{1}{x}$ does not exists (in the sense that it's value is not finite) then it's equivalent with proving $\forall M \in \mathbb{R}: \lim_{x \to 0^{+}} \frac{1}{x} \neq M$. This last statement is equivalent with
$$\forall M \in \mathbb{R}: \exists \varepsilon>0, \forall \delta>0, \exists x: 0<x-a<\delta \ \mbox{and} \left |\frac{1}{x}-M \right| \geq \varepsilon$$

To prove the above statement. Let $\epsilon = 1$ for example and try to find an $x$ such that the above statement holds.

Thanks Siron.

Let $\epsilon = 1$ so then we get:

$|\frac{1}{x} - M| > 1$

$|\frac{1}{x}| > 1 + |M| $ [by the triangle inequality]

$ x < \frac{1}{1 + |M|}$

So if $ x < \frac{1}{1 + |M|}$ then $|\frac{1}{x} - M| > 1$

So for $\delta = \frac{1}{1 + |M|}$ the limit as $x \to 0^{+}$ does not exist.

But this is only one $\delta$

How do you prove for all $\delta$?

In limit proofs, generally, how do you prove $|f(x) - L| > \epsilon$?

Do you have to prove for a SPECIFIC $\delta$ or ALL $\delta$?

Thanks!

 

[Siron]

Thanks Siron.

Let $\epsilon = 1$ so then we get:

$|\frac{1}{x} - M| > 1$

$|\frac{1}{x}| > 1 + |M| $ [by the triangle inequality]

$ x < \frac{1}{1 + |M|}$

So if $ x < \frac{1}{1 + |M|}$ then $|\frac{1}{x} - M| > 1$

So for $\delta = \frac{1}{1 + |M|}$ the limit as $x \to 0^{+}$ does not exist.

But this is only one $\delta$

How do you prove for all $\delta$?

In limit proofs, generally, how do you prove $|f(x) - L| > \epsilon$?

Do you have to prove for a SPECIFIC $\delta$ or ALL $\delta$?

Thanks!

Remark: note that I wrote $0<x-a<\delta$ in my previous post but this has to be $0<x<\delta$ of course.

It's true that we want to prove it $\forall \delta>0$. More precisely, we want to find $\forall \delta>0$ a suitable $x$ such that the statement holds.

Let $M \in \mathbb{R}$. Choose $\epsilon = 1$ and let $\delta>0$ be arbitrary. We'll translate the statement '$\forall \delta>0, \exists x$' as follows:

Let $N>0$ a real number such that $\delta > \frac{1}{N}$ (it's always possible to find such $\delta$) and suppose $x \in \left ]0,\frac{1}{N} \right[$. Clearly, $0<x<\frac{1}{N}$ by construction.

Knowing the above we get the following
$$\left|\frac{1}{x}-M \right | \geq \frac{1}{|x|}-|M| \geq N - |M|$$

Note that $N - |M|$ can be made greater than or equal to $1$. To see this, since $N>0$ we can set $N \geq |M|+\epsilon = |M|+1$ at the start of the proof.