limits: true or false: If limit for f(x) exists, then limit for |f(x)| exists.

[fresh]

a) if limit [FONT=&quot]of f[/FONT](x) as x approaches a exists, then limit of |f(x)| as x approaches a exists.

[FONT=&quot]I would say this is false
because although the limit of y=x as x approaches 0 exists, the limit does not exist for y= |x | as x approaches 0. [/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]b) if limit of |f(x)| as x approaches a exists, then limit of f(x) as x approaches a exists [/FONT]
[FONT=&quot]
I would say this is true. because any point on the absolute value function y= |x | where the limit does exist, (any point other than 0), a limit exists for the function y=x
[/FONT]
[FONT=&quot]
does my logic make sense? is there a better way to prove these statements are true or false algebraically? [/FONT]

 

[stapel]

a) if limit of f(x) as x approaches a exists, then limit of |f(x)| as x approaches a exists.

I would say this is false because although the limit of y=x as x approaches 0 exists, the limit does not exist for y= |x | as x approaches 0.

b) if limit of |f(x)| as x approaches a exists, then limit of f(x) as x approaches a exists

I would say this is true. because any point on the absolute value function y= |x | where the limit does exist, (any point other than 0), a limit exists for the function y=x

does my logic make sense? is there a better way to prove these statements are true or false algebraically?

Your "because" statements (your "proofs") seem to be restatements of your original "true" or "false" answer. That is, as I'm reading them, your answers amount to "yes/no, because it's true/not true". But what is your logical basis for thinking "true" or "false"?

For instance, if you feel that Statement (a) is false, what is your counter-example (that is, what is your proof)? If you feel that Statement (b) is true, then assume that the limit exists for the absolute value of the function, and then show that it exists for the original function.

 

[fresh]

Your "because" statements (your "proofs") seem to be restatements of your original "true" or "false" answer. That is, as I'm reading them, your answers amount to "yes/no, because it's true/not true". But what is your logical basis for thinking "true" or "false"?

For instance, if you feel that Statement (a) is false, what is your counter-example (that is, what is your proof)? If you feel that Statement (b) is true, then assume that the limit exists for the absolute value of the function, and then show that it exists for the original function.

a) limx =0
x>0

but lim|x| = DNE
x>0

so the statement is false

I can prove that this limit does not exist by taking the limit of [f(0+h)-f(0)]/h as h approaches 0 from the right and the left.

(|0+h| - |0|)/h further simplifies to |h|/h
limit of|h|/h as h approaches 0 from the right is 1 and limit of |h|/h as h approaches 0 from the left is -1
thus the overall limit of y=|x| does not exist because of the abrupt change in slope


b) lim|x|=1
x>1

limx =1
x>1

so the second statement is true.

but this is just one example where this is true. are there any examples where this is false? I don't think so, but I don't know how to prove it.

 

[ksdhart2]

a) limx =0
x>0

but lim|x| = DNE
x>0

so the statement is false

I can prove that this limit does not exist by taking the limit of [f(0+h)-f(0)]/h as h approaches 0 from the right and the left.

(|0+h| - |0|)/h further simplifies to |h|/h
limit of|h|/h as h approaches 0 from the right is 1 and limit of |h|/h as h approaches 0 from the left is -1
thus the overall limit of y=|x| does not exist because of the abrupt change in slope

This logic doesn't work at all. If you were trying to prove that |x| is not differentiable (i.e. the first derivative is undefined) at the point x = 0, this would be an excellent proof. However, that's not your goal. You're trying to find if the limit as x approaches 0 exists. Let's look at what happens as x approaches 0 from above:

• x = 1 ; |x| = 1
• x = 0.1 ; |x| = 0.1
• x = 0.01 ; |x| = 0.01
• x = 0.001 |x| = 0.001

Seems pretty clear to me that the values are closing in on 0. So what about as x approaches 0 from below?

• x = -1 ; |x| = 1
• x = -0.1 ; |x| = 0.1
• x = -0.01 ; |x| = 0.01
• x = -0.001 |x| = 0.001

Hmm... looks to me like it's still closing in on 0. This can be fairly easily proven by noting that, for any positive x, |x| = x. Thus, for the "approaches from above" limit, we can replace |x| by x and we know the limit exists and is 0. We can also see that, for any negative x, |x| = -x. So by the same principle, we can replace |x| by -x in the "approaches from below" limit. We know this limit exists and is also equal to 0. Both the left- and right-side limits exist and both are the same value, hence the overall limit exists and is 0.

What happens if you use this exact same logic for some arbitrary f(x)? When x approaches 0 from above, we know that limit must be equal to the limit of f(x). And when x approaches 0 from below, we know that limit must be equal to the limit of f(-x). Where does this lead you? Is the statement true? Can you prove why it is or why it isn't?

b) lim|x|=1
x>1

limx =1
x>1

so the second statement is true.

but this is just one example where this is true. are there any examples where this is false? I don't think so, but I don't know how to prove it.

This will play out very very similar to the first proof, except now you're given that the limit as x approaches 0 of f(|x|) = 0. What does that mean about f(x) and f(-x)? Is the statement true? Can you prove why it is or why it isn't?

 

[fresh]

This logic doesn't work at all. If you were trying to prove that |x| is not differentiable (i.e. the first derivative is undefined) at the point x = 0, this would be an excellent proof. However, that's not your goal. You're trying to find if the limit as x approaches 0 exists. Let's look at what happens as x approaches 0 from above:

• x = 1 ; |x| = 1
• x = 0.1 ; |x| = 0.1
• x = 0.01 ; |x| = 0.01
• x = 0.001 |x| = 0.001

Seems pretty clear to me that the values are closing in on 0. So what about as x approaches 0 from below?

• x = -1 ; |x| = 1
• x = -0.1 ; |x| = 0.1
• x = -0.01 ; |x| = 0.01
• x = -0.001 |x| = 0.001

Hmm... looks to me like it's still closing in on 0. This can be fairly easily proven by noting that, for any positive x, |x| = x. Thus, for the "approaches from above" limit, we can replace |x| by x and we know the limit exists and is 0. We can also see that, for any negative x, |x| = -x. So by the same principle, we can replace |x| by -x in the "approaches from below" limit. We know this limit exists and is also equal to 0. Both the left- and right-side limits exist and both are the same value, hence the overall limit exists and is 0.

What happens if you use this exact same logic for some arbitrary f(x)? When x approaches 0 from above, we know that limit must be equal to the limit of f(x). And when x approaches 0 from below, we know that limit must be equal to the limit of f(-x). Where does this lead you? Is the statement true? Can you prove why it is or why it isn't?



This will play out very very similar to the first proof, except now you're given that the limit as x approaches 0 of f(|x|) = 0. What does that mean about f(x) and f(-x)? Is the statement true? Can you prove why it is or why it isn't?

[FONT=&quot]if limit of f(x) as x approaches a exists, then[/FONT]
[FONT=&quot]lim_x→af(x) = L (a finite number)[/FONT]
[FONT=&quot]lim_x→a|f(x)| = |lim_x→af(x)| [/FONT]
[FONT=&quot] =|L|, a non negative finite number[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]So the statement "if limit of f(x) as x approaches a exists, then limit of |f(x)| as x approaches a exists" would be true? [/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]if limit of |f(x)| as x approaches a exists, then[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]lim_x→a|f(x)| = L where L is a non negative finite number[/FONT]
[FONT=&quot]|lim_x→af(x)|= L[/FONT]
[FONT=&quot]lim_x→af(x) = L or −L, a finite number.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]So the statement "if limit of |f(x)| as x approaches a exists, then limit of f(x) as x approaches a exists" is TRUE.[/FONT]

 

[ksdhart2]

if limit of f(x) as x approaches a exists, then
lim_x→af(x) = L (a finite number)
lim_x→a|f(x)| = |lim_x→af(x)|
=|L|, a non negative finite number

So the statement "if limit of f(x) as x approaches a exists, then limit of |f(x)| as x approaches a exists" would be true?

if limit of |f(x)| as x approaches a exists, then

lim_x→a|f(x)| = L where L is a non negative finite number
|lim_x→af(x)|= L
lim_x→af(x) = L or −L, a finite number.

So the statement "if limit of |f(x)| as x approaches a exists, then limit of f(x) as x approaches a exists" is TRUE.

It seems okay to me, but your instructor may want you to justify that the absolute value operator "distributes" over a limit like that. I imagine you could construct such a proof using the epsilon-delta definition of a limit. But I'd recommend asking your instructor if this suffices, as there's no sense in doing extra work if you don't have to.