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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution

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

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Science Advisor 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..
 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.

Recognitions:

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

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

 Even when I try to show that the limits are different either side of zero I keep getting divisions by zero.
 Recognitions: Science Advisor Ultimately, you can use the fact that cos is bounded , but x can be made indefinitely small. Have you tried a δ-ε proof?
 is it sometimes referred to as the squeeze theorem
 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.
 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.

 Quote by Zondrina 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=*FVarOp t.1-_**-.***Limit.limitvariable--.**Limit.direction--.**Limit.limitvariable2-.*Limit.limit2-.*Limit.direction2---.*--

Mentor
 Quote by quadreg 1. The problem statement, all variables and given/known data Prove algebraically that lim x->0 (x^2)*cos(1/ x^(1/3)) does not exist 2. Relevant equations lim x->0 (x^2)*cos(1/ x^(1/3)) 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.

Mentor
 Quote by quadreg 1. The problem statement, all variables and given/known data 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?

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

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

 Quote by clamtrox 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

Recognitions:
Homework Help
 Quote by clamtrox Because $x^{1/3}$is complex for $x<0$
No it's not 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

Recognitions:
 Quote by Mentallic No it's not 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
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 cant 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.