Find the following limits
$\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}$
Any hints/solutions how to approach this? I tried many ways, rationalization, taking out x, etc. But I still can't rid myself of the singularity. Thanks in advance.
Also another question.
Find the limit of $\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$
I worked up till here, after which I got stuck. I think I need to apply the squeeze theore, but I am not sure how to.
$\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2} = \lim_{x\to 0}\frac{-2\sin\frac{1}{2}(3x+x)\sin\frac{1}{2}(3x-x)}{x^2}=\lim_{x\to 0}\frac{-2\sin2x\sin x}{x^2}=\lim_{x\to 0}\frac{-2(2\sin x\cos x)\sin x}{x^2}=\lim_{x\to 0}\frac{-4\sin^2 x\cos x}{x^2}$
Solutions or hints will be appreciated. Thanks in advance! L'hospital's rule not allowed.