# Thread: Is this a false proof?

1. ## Is this a false proof?

Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Proof:

Let $f(x)=\frac{|x|}{x}.$ Assume, to the contrary, that $\lim_{x \to 0} f(x)$ exists. Then there exists a real number $L$ such that $\lim_{x \to 0} f(x) = L$. Let $\epsilon = 1$. Then there exists $\delta >0$ such that if $x$ is a real number satisfying $0<|x-0|<\delta$, then $|f(x)-L|<\epsilon$. We consider two cases.

Case 1. $L\geq0$. Consider $x=- \delta/2$. Then $|x|=\delta/2 <\delta$. However, $f(x)=f(-\delta/2) =(\delta/2)/(-\delta/2)=-1$. So $|f(x)-L|=|-1-L|=1+L > 1$, a contradiction.

Case 1. $L<0$. Consider $x=\delta/2$. Then $|x|=\delta/2 <\delta$. Also, $f(x)=f(\delta/2) =(\delta/2)/(\delta/2)=1$. So $|f(x)-L|=|1-L|=1-L > 1$, a contradiction.

Remarks: I am particularly troubled by the choice of $x$'s being contrary to the values of $L$. Intuitively, we know that the value of $x$ cannot be negative when $L+=\lim_{x \to 0^+}f(x)$ and $x$ cannot be positive when $L-=\lim_{x \to 0^-}f(x)$.

Please tell me where I am wrong.

2. Originally Posted by novice
Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Proof:

Let $f(x)=\frac{|x|}{x}.$ Assume, to the contrary, that $\lim_{x \to 0} f(x)$ exists. Then there exists a real number $L$ such that $\lim_{x \to 0} f(x) = L$. Let $\epsilon = 1$. Then there exists $\delta >0$ such that if $x$ is a real number satisfying $0<|x-0|<\delta$, then $|f(x)-L|<\epsilon$. We consider two cases.

Case 1. $L\geq0$. Consider $x=- \delta/2$. Then $|x|=\delta/2 <\delta$. However, $f(x)=f(-\delta/2) =(\delta/2)(-\delta/2)=-1$. So $|f(x)-L|=|-1-L|=1+L > 1$, a contradiction.

Case 1. $L<0$. Consider $x=\delta/2$. Then $|x|=\delta/2 <\delta$. Also, $f(x)=f(\delta/2) =(\delta/2)(\delta/2)=1$. So $|f(x)-L|=|1-L|=1-L > 1$, a contradiction.

Remarks: I am particularly troubled by the choice of $x$'s being contrary to the values of $L$. Intuitively, we know that the value of $x$ cannot be negative when $L+=\lim_{x \to 0^+}f(x)$ and $x$ cannot be positive when $L-=\lim_{x \to 0^-}f(x)$.

Please tell me where I am wrong.
Proof seems valid to me, although you made a typo saying multiplication

$f(-\delta/2) =(\delta/2)(-\delta/2)$

instead of division with absolute value.

What you're saying is that in any open ball centered at 0, we can find a value of f(x) that is -1, and a value of f(x) that is 1. So we will never be able to get |f(x)-L| bounded for any epsilon less than or equal to 1. (If epsilon is greater than 1, then we can always choose L = 0 and the bound holds.)

Another approach is simply to say the right- and left-hand limits exist and are not equal.

3. Is'nt $x$ a positive value when $L \geq 0$ and otherwise?

4. Originally Posted by novice
Is'nt $x$ a positive value when $L \geq 0$ and otherwise?
I don't know what you mean. L does not determine x. x is an indeterminate and takes on all values in the domain. If we restrict x to the set (-delta, delta) \ {0}, then we consider all possible x with corresponding f(x) in that set. There are only two possible f(x) because if x < 0 then f(x) = -1, and if x > 0 then f(x) = 1.

There does not exist an L satisfying the definition of limit, because the limit does not exist. But L = 0 gives the "best" candidate because it straddles the two possible values of f(x) and thus epsilon can be chosen smaller than for any other value of L. For example if we say candidate L = 0.1 then for epsilon = 1.1 the epsilon-delta condition would fail. But epsilon = 1.1 holds for candidate L = 0.

5. Originally Posted by novice
Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Proof:

Let $f(x)=\frac{|x|}{x}.$ Assume, to the contrary, that $\lim_{x \to 0} f(x)$ exists. Then there exists a real number $L$ such that $\lim_{x \to 0} f(x) = L$. Let $\epsilon = 1$. Then there exists $\delta >0$ such that if $x$ is a real number satisfying $0<|x-0|<\delta$, then $|f(x)-L|<\epsilon$. We consider two cases.

Case 1. $L\geq0$. Consider $x=- \delta/2$. Then $|x|=\delta/2 <\delta$. However, $f(x)=f(-\delta/2) =(\delta/2)/(-\delta/2)=-1$. So $|f(x)-L|=|-1-L|=1+L > 1$, a contradiction.

Case 1. $L<0$. Consider $x=\delta/2$. Then $|x|=\delta/2 <\delta$. Also, $f(x)=f(\delta/2) =(\delta/2)/(\delta/2)=1$. So $|f(x)-L|=|1-L|=1-L > 1$, a contradiction.

Remarks: I am particularly troubled by the choice of $x$'s being contrary to the values of $L$. Intuitively, we know that the value of $x$ cannot be negative when $L+=\lim_{x \to 0^+}f(x)$ and $x$ cannot be positive when $L-=\lim_{x \to 0^-}f(x)$.

Please tell me where I am wrong.
Sorry, but what's the contradiction in both cases?!
In the 1st you assumed L>=0 and at last you obtained L+1>1 which means L>0.
In the 2nd you got L<0 and at last you recived 1-L>1 which means L<0.
Thanks

6. Originally Posted by Mathelogician
Sorry, but what's the contradiction in both cases?!
In the 1st you assumed L>=0 and at last you obtained L+1>1 which means L>0.
In the 2nd you got L<0 and at last you recived 1-L>1 which means L<0.
Thanks
The contradiction is that we assumed the value is less than epsilon = 1.

However your comment caused me to look closer and I noticed that where it's written 1 + L > 1 what it should be is 1 + L is greater than or equal to 1.

7. I would probably proceed by showing that the RH limit and the LH limit do not agree.

8. The function $|x|$ is V-shaped with lines at 45 degrees approaching each other at the origin.

The function $\frac{|x|}{x}$ is a step function,

the function does not contain $x=0$ as part of the domain as the denominator is x.

There is a discontinuity at the y-axis.

$f(x)=\frac{|x|}{x}=-1$ for x negative.

$f(x)=\frac{|x|}{x}=1$ for x positive.

Even L'Hopital's rule gives 2 seperate limits for x to the left and right of the y-axis.

9. Originally Posted by novice
Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Proof:

Let $f(x)=\frac{|x|}{x}.$ Assume, to the contrary, that $\lim_{x \to 0} f(x)$ exists. Then there exists a real number $L$ such that $\lim_{x \to 0} f(x) = L$. Let $\epsilon = 1$. Then there exists $\delta >0$ such that if $x$ is a real number satisfying $0<|x-0|<\delta$, then $|f(x)-L|<\epsilon$.

***** If a limit L exists, then f(x) approaches that limit, hence f(x)-L must be at least <1 at some point*******

We consider two cases.

Case 1. $L\geq0$. Consider $x=- \delta/2$.

******But if $L\ge0,$ x must be $\ge0$******

Then $|x|=\delta/2 <\delta$. However, $f(x)=f(-\delta/2) =(\delta/2)/(-\delta/2)=-1$. So $|f(x)-L|=|-1-L|=1+L > 1$, a contradiction.

****** $f(x)=1, |f(x)-L|=|1-L|$********

Case 2. $L<0$. Consider $x=\delta/2$.

****** $x=-\frac{\delta}{2}$*******

Then $|x|=\delta/2 <\delta$. Also, $f(x)=f(\delta/2) =(\delta/2)/(\delta/2)=1$.

****** $f(x)=-1$*******

So $|f(x)-L|=|1-L|=1-L > 1$, a contradiction.

****** $|f(x)-L|=|-1-L|******$

Remarks: I am particularly troubled by the choice of $x$'s being contrary to the values of $L$. Intuitively, we know that the value of $x$ cannot be negative when $L+=\lim_{x \to 0^+}f(x)$ and $x$ cannot be positive when $L-=\lim_{x \to 0^-}f(x)$.

Please tell me where I am wrong.
Looking at it another way,

for case 1... $f(x)-L=0$ only if $L=1.$

for case 2... $f(x)-L=0$ only if $L=-1.$