Find the following limit $\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}$ and $\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$

Question

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.

Answer 1

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 $(a^3-b^3)=(a-b)(a^2+ab+b^2)$.

Answer 2

For the first one, you need to rationalize. The formula for $a^3-b^3$ yields:

$$(\sqrt[3]{1+x}-1)(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1)= (1+x)-1 \,.$$

Alternately, what you have there is the definition of the derivative of $\sqrt[3]{1+x}$ at $x=0$.

For the second one, at this step you are done:

$$\lim_{x\to 0}\frac{-2\sin2x*\sin x}{x^2} \,.$$

Just use

$$\lim_{x\to 0}\frac{\sin2x}{2x}=\lim_{x\to 0}\frac{\sin x}{x}=1$$

Answer 3

Like N.S. said, looking this limit as derivative is a way to solve. You could also do $u=x+1$ to simplify your expression and consider $f(u)=u^{1/3}$.

$$u=x+1\rightarrow \lim_{u\rightarrow 1} \frac{u^{1/3}-1}{u-1}=\lim_{u\rightarrow 1} \frac{f(u)-f(1)}{u-1}=f'(1)$$

But $f'(u) = \frac{1}{3}u^{-2/3}$, then $f'(1) = \frac{1}{3}\cdot 1^{-2/3}=\frac{1}{3}$.