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

[quadreg]

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.

 

[Bacle2]

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..

 

[clamtrox]

L'Hopital does not apply if the limit is not in the specific form required by it...

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.

 

[Bacle2]

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

 

[quadreg]

Even when I try to show that the limits are different either side of zero I keep getting divisions by zero.

 

[Bacle2]

Ultimately, you can use the fact that cos is bounded , but x can be made indefinitely

small. Have you tried a δ-ε proof?

 

[quadreg]

is it sometimes referred to as the squeeze theorem

 

[clamtrox]

So this is a stupid question but how do you know the limit does not exist? Perhaps you could use that direction of reasoning in your proof.

It's not very difficult to show either that the limit
$$\lim_{x\rightarrow 0^-}x^2 \cos(x^{-1/3}) \sim \lim_{x\rightarrow 0} x^2 \exp(|x|^{-1/3})$$ which does not have limit 0.

 

[STEMucator]

You're over thinking this question way too much. Recall that $\forall$x$\inℝ$ :

$|cos(x)| ≤ 1$

Which translates into :

$-1 ≤ cos(x) ≤ 1$

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

 

[quadreg]

You're over thinking this question way too much. Recall that $\forall$x$\inℝ$ :

$|cos(x)| ≤ 1$

Which translates into :

$-1 ≤ cos(x) ≤ 1$

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

so I get

−1≤cos($\frac{1}{\sqrt[3]{x}}$)≤1

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

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---.*--

 

[SammyS]

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.

Well, assuming that the cube root is a real function, this limit does exist, & it is zero.

 

[Mark44]

Homework Statement

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

Are you sure you wrote down the problem correctly?

 

[Mentallic]

I'm curious with what's happening in Wolfram Alpha's answer. Why is it claiming that $$\lim_{x\to 0^-}f(x)=\tilde{\infty}$$

 

[clamtrox]

I'm curious with what's happening in Wolfram Alpha's answer. Why is it claiming that $$\lim_{x\to 0^-}f(x)=\tilde{\infty}$$

Because $x^{1/3}$is complex for $x<0$ 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.

 

[quadreg]

Because $x^{1/3}$is complex for $x<0$ 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 $x^{1/3}$, include all ℝ, you can take the cube root of a negative can't you.

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

 

[Mentallic]

Because $x^{1/3}$is complex for $x<0$

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 $re^{i\pi / 3}$, it'll use that as opposed to our expected $re^{i\pi}$.

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

 

[uart]

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 $re^{i\pi / 3}$, it'll use that as opposed to our expected $re^{i\pi}$.

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 $1 + i \sqrt{3}$ for (-8)^(1/3).

Pylab gives an error message ('negative can't be raised to a fractional power') if you try (-8.0)**(1/3.0) but returns $1 + \sqrt{3}j$ for (-8.0 + 0.0j)^(1/3.0). In the misc special functions it does include one for real cube root, scipy.special.cbrt(-8.0) returns -2.0.

 

[quadreg]

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.

 

[Bacle2]

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.