Anything missing or redundant about this one-sided limit proof?

[mcastillo356]

Homework Statement

Prove $\lim{(x^{2/3})}$ when $x\rightarrow{0^{+}}$ is 0

Relevant Equations

$\forall{\epsilon>0}$, find $\delta>0$ such that $0<x<\delta\Rightarrow{|f(x)|<\epsilon}$

Hi, PF

In a Spanish math forum I got this proof of a right hand limit:

"For a generic $\epsilon>0$, in case the inequality is met, we have the following: $|x^{2/3}|<\epsilon\Rightarrow{|x|^{2/3}}\Rightarrow{|x|<\epsilon^{3/2}}$. Therein lies the condition. If $x>0$, then $|x|=x$; therefore, if the following holds: $0<x<\epsilon^{3/2}\Rightarrow{|f(x)|<\epsilon}$, eventually, we can state: $\forall{\epsilon>0}\;\exists{\delta>0}$ s.t. $0<x<\delta\Rightarrow{|f(x)|<\epsilon}$. In conclusion, the $\delta$ sought is epsilon elevated to three means, $\delta=\epsilon^{3/2}$."

What is your opinion? It's right, yes, but... Something tells me it shall be improved.

Love

I'm going to click, no preview.

 

[Mark44]

In conclusion, the δ sought is epsilon elevated to three means, δ=ϵ^3/2."

Epsilon raised to the three-halves power. Since the purported limit is 0, it would be nice, but not essential to start with $|x^{2/3} - 0| < \epsilon$, but that's a very minor nit. Otherwise, I don't see anything wrong with the proof.

 

[mcastillo356]

Proof of left handed limit for $x^{3/2}$, when $x\rightarrow{0^{-}}$

For a generic $\epsilon>0$, in case the inequality mets (or inequation? Is it an algebraic inequality?), we have the following: $|x^{2/3}|<\epsilon\Rightarrow{|x|^{2/3}< \epsilon}\Rightarrow{|x|<\epsilon^{3/2}}$. Therein lies the condition. If $x<0$, then $|x|=-x$; therefore, if the following holds: $-\epsilon^{3/2}<x<0\Rightarrow{|f(x)|<\epsilon}$, eventually, we can state, $\forall{\epsilon>0}\;\exists{\delta>0}$ s.t. $-\delta<x<0\Rightarrow{|f(x)|<\epsilon}$. In conclusion, the $\delta$ sought is epsilon raised to the three-halves power, $\delta=\epsilon^{3/2}$.

Right?

 

[Mark44]

Proof of left handed limit for $x^{3/2}$, when $x\rightarrow{0^{-}}$

For a generic $\epsilon>0$, in case the inequality mets (or inequation? Is it an algebraic inequality?), we have the following: $|x^{2/3}|<\epsilon\Rightarrow{|x|^{2/3}< \epsilon}\Rightarrow{|x|<\epsilon^{3/2}}$. Therein lies the condition. If $x<0$, then $|x|=-x$; therefore, if the following holds: $-\epsilon^{3/2}<x<0\Rightarrow{|f(x)|<\epsilon}$, eventually, we can state, $\forall{\epsilon>0}\;\exists{\delta>0}$ s.t. $-\delta<x<0\Rightarrow{|f(x)|<\epsilon}$. In conclusion, the $\delta$ sought is epsilon raised to the three-halves power, $\delta=\epsilon^{3/2}$.

Right?

No.
$x^{3/2}$ isn't defined for x < 0 if you're limited to real output values.

Think about it -- if x = -0.1, say, then you have to evaluate either $\sqrt{-.1^3}$, or $(\sqrt{-.1})^3$, neither of which is real.

The reason for the right-hand limit of the first post is precisely because the function isn't defined for negative x, so the left-hand limit doesn't exist, either.

 

[mcastillo356]

I disagree. We deal with absolute values

 

[Office_Shredder]

I disagree. We deal with absolute values

Only a sith deals in absolutes.

But seriously, what does this mean? Is your function actually $|x|^{3/2}$?

 

[Mark44]

I disagree. We deal with absolute values

That's not the limit you wrote in post #3:

Proof of left handed limit for $x^{3/2}$, when $x\rightarrow{0^{-}}$

Again, $f(x) = x^{3/2}$ is defined (as a real valued function) only for $x \ge 0$.

 

[mcastillo356]

I meant $f(x)=x^{2/3}$

 

[mcastillo356]

Sorry, I will try to mend it.

 

[Office_Shredder]

I think the proof is fine then.

 

[mcastillo356]

Hi, PF

No.
$x^{3/2}$ isn't defined for x < 0 if you're limited to real output values.

Think about it -- if x = -0.1, say, then you have to evaluate either $\sqrt{-.1^3}$, or $(\sqrt{-.1})^3$, neither of which is real.

The reason for the right-hand limit of the first post is precisely because the function isn't defined for negative x, so the left-hand limit doesn't exist, either.

$\lim{(x^{3/2})}$ when $x\rightarrow{0^{+}}$ is $0$

Proof

$\forall{\epsilon>0}\;\exists{\delta>0}$ s.t. $0<x<\delta\Rightarrow{|f(x)|<\epsilon}$

Pick a $\delta=\epsilon^{2/3}$. This way, $f(x)$ can be as closer to $0$ as desired, getting $x$ close enough to $0$, $0<x<\delta=\epsilon^{2/3}$

This right hand limit equals $f(0)=0$, but it will never be continous, as @Mark44 suggests, and the graph shows:



Right? I've wrote not checking.

 

[Mark44]

Hi, PF
$\lim{(x^{3/2})}$ when $x\rightarrow{0^{+}}$ is $0$

Proof

$\forall{\epsilon>0}\;\exists{\delta>0}$ s.t. $0<x<\delta\Rightarrow{|f(x)|<\epsilon}$

Pick a $\delta=\epsilon^{2/3}$. This way, $f(x)$ can be as closer to $0$ as desired, getting $x$ close enough to $0$, $0<x<\delta=\epsilon^{2/3}$

You should show explicitly how $\delta=\epsilon^{2/3}$ implies that $|x^{3/2} - 0 | < \epsilon$.

This right hand limit equals $f(0)=0$, but it will never be continous, as @Mark44 suggests, and the graph shows:

No, that's not what I suggested. The function $f(x) = x^{3/2}$ is continuous on its domain, $[0, \infty)$, and is right-continuous at 0.

View attachment 294897

Right? I've wrote not checking.

 

[Office_Shredder]

Re reading this thread, I think there is actually a piece that is not spelled out specifically enough. As written, you can basically replace $f(x)=x^{2/3}$ with any invertible function, and it's not super obvious which step stops working. Do you know?

 

[mcastillo356]

Re reading this thread, I think there is actually a piece that is not spelled out specifically enough. As written, you can basically replace $f(x)=x^{2/3}$ with any invertible function, and it's not super obvious which step stops working. Do you know?

Sorry, can you explain further?

 

[Office_Shredder]

Sorry, can you explain further?

Let $f$ be an invertible function with $f(0)=0$. Then for any $\epsilon>0$, let $\delta=f^{-1}(\epsilon)$. Then $0<x< f^{-1}(\epsilon)$ implies $f(x)<\epsilon$. Hence $\lim_{x\to 0^+} f(x)=0$.

Do you see the error in this proof for an arbitrary function? I just copied what you did for $x^{2/3}$

 

[mcastillo356]

Let me see... It only works because it's fractional, even, and positive power function. No. Counterexample: absolute value function for real numbers; square root function. Haven't checked, give me some time

 

[Office_Shredder]

Let me see... It only works because it's fractional, even, and positive power function. No. Counterexample: absolute value function for real numbers; square root function. Haven't checked, give me some time

It might help to think of an invertible function $f:[0,\infty) \to \mathbb{R}$ which is *not* continuous at 0 (this is not super hard, but not a totally trivial thing to do)

 

[mcastillo356]

Let $f$ be an invertible function with $f(0)=0$. Then for any $\epsilon>0$, let $\delta=f^{-1}(\epsilon)$. Then $0<x< f^{-1}(\epsilon)$ implies $f(x)<\epsilon$. Hence $\lim_{x\to 0^+} f(x)=0$.

Do you see the error in this proof for an arbitrary function? I just copied what you did for $x^{2/3}$

But it is an increasing function. $f(x)<\epsilon\;\forall{\epsilon>0}$

 

[haruspex]

But it is an increasing function.

But you did not note the dependence on that in your proof, let alone prove it.

 

[mcastillo356]

But you did not note the dependence on that in your proof, let alone prove it.

Incisive remark. Definitely, there was something missing. I'm working on it.