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

  • Thread starter mcastillo356
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    Limit Proof
In summary: Actually, it doesn't matter if the function is increasing or not. The error lies in assuming that for any ##\epsilon>0##, there exists a single value of ##\delta## that works for all ##x## in the interval ##(0, \delta)##. This is not true for all functions, as the counterexamples you mentioned (absolute value and square root functions) show.
  • #1

mcastillo356

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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.🤞
 
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  • #2
mcastillo356 said:
In conclusion, the δ sought is epsilon elevated to three means, δ=ϵ3/2."
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.
 
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  • #3
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?
 
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  • #4
mcastillo356 said:
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.
 
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  • #5
I disagree. We deal with absolute values
 
  • #6
mcastillo356 said:
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}##?
 
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  • #7
mcastillo356 said:
I disagree. We deal with absolute values
That's not the limit you wrote in post #3:
mcastillo356 said:
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##.
 
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  • #8
:doh:
I meant ##f(x)=x^{2/3}##
 
  • #9
Sorry, I will try to mend it.
 
  • #10
I think the proof is fine then.
 
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  • #11
Hi, PF

Mark44 said:
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:

geogebra-export (3).png

Right? I've wrote not checking.
 
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  • #12
mcastillo356 said:
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##.
mcastillo356 said:
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.
mcastillo356 said:
View attachment 294897

Right? I've wrote not checking.
 
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  • #13
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?
 
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  • #14
Office_Shredder said:
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?
 
  • #15
mcastillo356 said:
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}##
 
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  • #16
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
 
  • #17
mcastillo356 said:
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)
 
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  • #18
Office_Shredder said:
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}##:confused:
 
  • #19
mcastillo356 said:
But it is an increasing function.
But you did not note the dependence on that in your proof, let alone prove it.
 
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  • #20
haruspex said:
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.
 

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