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.


*

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

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

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.
 A: You don't end up with $0/0$ in the first case: in
$$
\frac{(x+1)-1}{x(\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1)}
$$
the factor $x$ cancels out and you get $1/3$.
For the second one, recall the sum-to-product formula:
$$
\cos3x-\cos x=-2\sin2x\sin x
$$
A: As regards 2), note that
$$\frac{\cos(3x) - \cos(x)}{x^2}=\frac{1 - \cos x}{x^2}-9\cdot\frac{1 - \cos (3x)}{(3x)^2}.$$
Hence
$$\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}=
\lim_{x \to 0}\frac{1 - \cos x}{x^2}-9\cdot\lim_{x \to 0}\frac{1 - \cos (3x)}{(3x)^2}\\=(1-9)\lim_{x \to 0}\frac{1 - \cos x}{x^2}
=-8\lim_{x \to 0}\frac{1 - \cos^2 x}{x^2(1+\cos x)}=-\frac{8}{2}\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^2=-4.$$
A: $$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}=\lim_{x \to 0} \frac{(\sqrt[3]{1+x}-1)(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1)}{x(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1)}=\lim_{x \to 0} \frac{1+x-1}{x(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1)}=\lim_{x \to 0}\frac{1}{\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1}=\frac{1}{3}$$
A: Q1 - good approach. But take $a=\sqrt[3]{1+x}$, $b=1$. See what happens. Not $\frac{0}{0}$.
Q2 - similarly - something should cancel and $\frac{0}{0}$ should vanish.
A: Hint:)
For the first use following formula
$$a^3-b^3=(a-b)(a^2+ab+b)$$
for denominator $x=(1+x)-1=\sqrt[3]{1+x}^3-1^3$ and  for the second use correct formula
$$\cos 3x-\cos x =-2\sin 2x \sin x$$
A: *

*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 $$\frac{\cos3x-\cos x}{x^2}=\frac{(1-\frac{9x^2}{2}+\frac{81x^4}{4}-\ldots)-(1-\frac{x^2}{2}+\frac{x^4}{4}-\ldots)}{x^2}=\frac{\frac{-9x^2}{2}+\frac{x^2}{2}+\frac{81x^4}{4}-\frac{x^4}{4}+\ldots}{x^2}=-4+20x^2+\ldots\implies\lim_{x\to0}\frac{\cos3x-\cos x}{x^2}=-4$$
A: 


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

$$
\lim _{x\rightarrow 0}\dfrac{\cos \left( ax\right) -\cos \left( bx\right) }{\left( x\right) ^{2}} =\dfrac{b^{2}-a^{2}}{2}
$$
Therefore,
$$
\lim _{x\rightarrow 0}\dfrac{\cos 3x-\cos x}{x^{2}}=\dfrac{\left( 1\right) ^{2}-\left( 3\right) ^{2}}{2}=-4
$$
