- #1

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## Homework Statement

Prove algebraically that lim x->0 (x^2)*cos(1/ x^(1/3)) does not exist

## Homework Equations

lim x->0 (x^2)*cos(1/ x^(1/3))

## The Attempt at a Solution

I know the limit does not exist but can't prove it algebraically.

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- Thread starter quadreg
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- #1

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Prove algebraically that lim x->0 (x^2)*cos(1/ x^(1/3)) does not exist

lim x->0 (x^2)*cos(1/ x^(1/3))

I know the limit does not exist but can't prove it algebraically.

- #2

Bacle2

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Have you tried something like, say, L'Hopital's rule?

Edit: It seems from the graph:

http://www.meta-calculator.com/online/

that the limit may exist. Of course, this is not a proof. But the

derivative near zero does not blow up. Moreover, the whole expression is bounded by +/ 1.

You can also consider what happens with different sequences of points as you approach 0.

I guess points where there may be trouble are multiples of Pi, say , 1/x^{1/3} =n*Pi..

Edit: It seems from the graph:

http://www.meta-calculator.com/online/

that the limit may exist. Of course, this is not a proof. But the

derivative near zero does not blow up. Moreover, the whole expression is bounded by +/ 1.

You can also consider what happens with different sequences of points as you approach 0.

I guess points where there may be trouble are multiples of Pi, say , 1/x^{1/3} =n*Pi..

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- #3

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It's clear that when you approach zero from above, the limit is zero. I guess you need to show that approaching from below, it's not.

- #4

Bacle2

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You can rewrite the expression, "massage it" , to put it in a form that fits L'Hopital.

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- #6

Bacle2

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small. Have you tried a δ-ε proof?

- #7

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is it sometimes referred to as the squeeze theorem

- #8

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It's not very difficult to show either that the limit

[tex] \lim_{x\rightarrow 0^-}x^2 \cos(x^{-1/3}) \sim \lim_{x\rightarrow 0} x^2 \exp(|x|^{-1/3}) [/tex] which does not have limit 0.

- #9

STEMucator

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[itex]|cos(x)| ≤ 1[/itex]

Which translates into :

[itex]-1 ≤ cos(x) ≤ 1[/itex]

Then simply remember how to manipulate this inequality and this problem will become all too easy for you.

- #10

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[itex]|cos(x)| ≤ 1[/itex]

Which translates into :

[itex]-1 ≤ cos(x) ≤ 1[/itex]

Then simply remember how to manipulate this inequality and this problem will become all too easy for you.

so I get

−1≤cos([itex]\frac{1}{\sqrt[3]{x}}[/itex])≤1

−x[itex]^{2}[/itex]≤x[itex]^{2}[/itex]cos([itex]\frac{1}{\sqrt[3]{x}}[/itex])≤x[itex]^{2}[/itex]

so as x -> 0 bounds -> 0, so the limit is 0.

however wolfram alpha tells me otherwise http://www.wolframalpha.com/input/?i=limits&a=*C.limits-_*Calculator.dflt-&f2=%28x^2%29*cos%281%2F+x^%281%2F3%29%29+&f=Limit.limitfunction_%28x^2%29*cos%281%2F+x^%281%2F3%29%29+&f3=0&x=2&y=3&f=Limit.limit_0&a=*FVarOpt.1-_**-.***Limit.limitvariable--.**Limit.direction--.**Limit.limitvariable2-.*Limit.limit2-.*Limit.direction2---.*--

- #11

SammyS

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Well, assuming that the cube root is a real function, this limit## Homework Statement

Prove algebraically that lim x->0 (x^2)*cos(1/ x^(1/3)) does not exist

## Homework Equations

lim x->0 (x^2)*cos(1/ x^(1/3))

## The Attempt at a Solution

I know the limit does not exist but can't prove it algebraically.

- #12

Mark44

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Are you sure you wrote down the problem correctly?## Homework Statement

Prove algebraically that lim x->0 (x^2)*cos(1/ x^(1/3)) does not exist

- #13

Mentallic

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- #14

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Because [itex] x^{1/3} [/itex]is complex for [itex]x<0 [/itex] and cosine of a complex number behaves very differently from cosine of a real number. In particular, it grows without bound if the argument has a large imaginary part.

- #15

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Because [itex] x^{1/3} [/itex]is complex for [itex]x<0 [/itex] and cosine of a complex number behaves very differently from cosine of a real number. In particular, it grows without bound if the argument has a large imaginary part.

Isn't the domain of [itex] x^{1/3} [/itex], include all ℝ, you can take the cube root of a negative can't you.

eg. [itex]\sqrt[3]{-125}[/itex]=-5

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- #16

Mentallic

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No it's not :tongue: The cube root of a negative number is still negative.Because [itex] x^{1/3} [/itex]is complex for [itex]x<0 [/itex]

But I think I see now what it's doing. For the cube root of the negative values, since the cube root can take complex values as well, Wolfram Alpha has probably been programmed to take the complex value with the smallest argument, and since this occurs at [itex]re^{i\pi / 3}[/itex], it'll use that as opposed to our expected [itex]re^{i\pi}[/itex].

I don't know about you, but I would call that a bug

- #17

uart

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No it's not :tongue: The cube root of a negative number is still negative.

But I think I see now what it's doing. For the cube root of the negative values, since the cube root can take complex values as well, Wolfram Alpha has probably been programmed to take the complex value with the smallest argument, and since this occurs at [itex]re^{i\pi / 3}[/itex], it'll use that as opposed to our expected [itex]re^{i\pi}[/itex].

I don't know about you, but I would call that a bug

It's an interesting problem. So simple if you use the real cube root, but such an interesting twist if you use the principle value of the complex exponentiation.

A lot of software uses this convention for (1/3)rd power of negative reals. Matlab and Octave both return [itex]1 + i \sqrt{3}[/itex] for (-8)^(1/3).

Pylab gives an error message ('

- #18

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Out of curiosity is it possible for there to be more than one limit, given there's more than one root, whether the limit exists or not.

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- #19

Bacle2

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is it sometimes referred to as the squeeze theorem

Only when the endpoints are both zero, otherwise you cannot squeeze precisely-enough between, say -2 and 2.

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