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

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.

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Expand into Taylor series upto power $3$ and cancel out the relevant terms. –  gt6989b Sep 24 '12 at 3:16
Haven't learnt Taylor's series either... :( –  Yellow Skies Sep 24 '12 at 3:20

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

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I'm sorry I forgot to post that I can't use the l'Hopital rule.Technically, I am not supposed to have learnt it yet. –  Yellow Skies Sep 24 '12 at 3:14
@TayBoonSiang: I wondered about that. Here’s an acceptable solution for your second limit. –  Brian M. Scott Sep 24 '12 at 3:20
And thank you!! –  Yellow Skies Sep 24 '12 at 3:30

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

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

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Thanks a bunch! really appreciate all the help. –  Yellow Skies Sep 24 '12 at 3:31

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

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Since $\cos(x) \sim -x^2/2$ if $x \to 0$, the second argument of your second limit is $\frac{-9x^2/2 + x^2/2}{x^2}$ which evaluates to $-4$; as for your first limit, since $(x+1)^{a} -1 \sim ax$ if $a > 0$ and $x\to 0$ you get that the argument is $(1/3)x /x$ which evaluates to $1/3$.

Note that $\sim$ means asymptotic to. Also, you might say that these are the first terms of the Taylor series of the functions in your limit, but those asymptotics can be proven without Taylor series.

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