<s>[INST] Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.


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*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$

*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$

I have tried to "rationalise" Q1 by using the identity $a^3-b^3=(a-b)(a^2+ab+b)$ but I still end up with a case of $\frac{0}{0}$.
For Q2, I tried to apply the formula and express $\cos 3x-\cos x =2\cos 2x \cos x$ which didn't help.
I would really be grateful for some advice on how I can proceed with the evaluating.
 [/INST] *

*We use the binomial theorem: $$\frac{\sqrt[3]{1+x}-1}{x}=\frac{(1+x)^{1/3}-1}{x}=\frac{(1+\frac1{3}x+\frac1{9}x^2+\ldots)-1}{x}=\frac{(\frac1{3}x+\frac1{9}x^2+\ldots)}{x}=(\frac1{3}+\frac1{9}x+\ldots)\implies\lim_{x\to0}\frac{\sqrt[3]{1+x}-1}{x}=\frac1{3}$$


2.Again, we use series expansion for $\cos$ function. We get $$\frac{\cos3x-\cos x}{x^2}=\frac{(1-\frac{9x^2}{2}+\frac{81x^4}{4}-\ldots)-(1-\frac{x^2}{2}+\frac{x^4}{4}-\ldots)}{x^2}=\frac{\frac{-9x^2}{2}+\frac{x^2}{2}+\frac{81x^4}{4}-\frac{x^4}{4}+\ldots}{x^2}=-4+20x^2+\ldots\implies\lim_{x\to0}\frac{\cos3x-\cos x}{x^2}=-4$$
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