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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.
so I getYou're over thinking this question way too much. Recall that [itex]\forall[/itex]x[itex]\inℝ[/itex] :
[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.
Well, assuming that the cube root is a real function, this limit does exist, & it is zero.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.
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
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.I'm curious with what's happening in Wolfram Alpha's answer. Why is it claiming that [tex]\lim_{x\to 0^-}f(x)=\tilde{\infty}[/tex]
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.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.
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]
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.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
Only when the endpoints are both zero, otherwise you cannot squeeze precisely-enough between, say -2 and 2.is it sometimes referred to as the squeeze theorem