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.
Best Answer
Revised to avoid l’Hospital’s rule:
Your second one can be finished off like this:
$$\begin{align*} \lim_{x\to 0}\frac{-2\sin 2x\sin x}{x^2}&=-2\left(\lim_{x\to 0}\frac{\sin 2x}x\right)\left(\lim_{x\to 0}\frac{\sin x}x\right)\\ &=-4\left(\lim_{x\to 0}\frac{\sin 2x}{2x}\right)\cdot1\\ &=-4\;. \end{align*}$$
Try multiplying the fraction in your first limit by
$$\frac{(1+x)^{2/3}+(1+x)^{1/3}+1}{(1+x)^{2/3}+(1+x)^{1/3}+1}$$
and making use of the identity $(a^3-b^3)=(a-b)(a^2+ab+b^2)$.