Recent content by quadreg

Rsl.

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.

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

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...

is it sometimes referred to as the squeeze theorem

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

Spiritual things do not exist.

Ohh, that makes sense. 3x^2 is always >=0 and (3^x)ln3 is always >=0, so f'(x)>0. Thankyou.

Yes, I tried solving for critical points, but i get f'(x)=3x^2+(3^x)ln3 and can't solve for x where f'(x) = 0

Two x values give the same y value. Are you suggesting I use the horizontal line test?

Homework Statement Prove f(x) = x^3 + 3^x is a one-to-one function. 2. The attempt at a solution Sum of one-to-one functions is a one-to-one function (I think/dont know how to prove). x^3 is one-to-one, 3^x is one-to-one, thus f(x) is one-to-one. Surely there's a more rigorous proof.