Easy way to compute this limit How to easily compute the following limit without l'Hospital : $$\lim_{x\to 0} \frac{\cos(x)+\cos(2x)-2}{\cos(3x)+\sin(x)-1}$$
Ps: Is it possible to use the derivative of a function ?
 A: Given limit
$$=\frac{\frac{\cos x-1}{x}+\frac{\cos 2x-1}{x}}{\frac{\cos 3x-1}{x}+\frac{\sin x}{x}}$$
Denominator goes to non-zero finite limit and numerator goes to $0$, using the fact that $$\lim_{x\to 0}\frac{\cos ax-1}{x}=\lim_{x\to 0}\frac{\cos ax - \cos (a\times 0)}{x-0}$$
$$=\frac{d}{dx}(\cos ax)|_0$$
$$=0$$
A: Besides L'Hospital's Rule, there is also one way to prove. Know that:
$$\lim_{x\rightarrow0} \dfrac{\cos ax-1}{x}=\lim_{x\rightarrow0} \dfrac{\left(\cos ax-1\right)\left(\cos ax+1\right)}{x\left(\cos ax+1\right)}=\lim_{x\rightarrow0} \dfrac{\cos^2 ax-1}{x\left(\cos ax+1\right)}=\lim_{x\rightarrow0} \dfrac{-\sin^2 ax}{x\left(\cos ax+1\right)}\\=-a\lim_{x\rightarrow0} \dfrac{\sin ax}{ax}\times\lim_{x\rightarrow0} \dfrac{\sin ax}{\cos ax+1}=0$$
Then, divide both side by $x$ in the limit, we get:
$$\lim_{x\rightarrow0} \dfrac{\cos x+\cos 2x-2}{\cos 3x+\sin x-1}=\lim_{x\rightarrow0} \dfrac{\frac{\cos x-1}{x}+\frac{\cos 2x-1}{x}}{\frac{\cos 3x-1}{x}+\frac{\sin x}{x}}=\dfrac{0+0}{0+1}=0$$
A: Write out the first few terms of the power series expansions of the numerator and denominator. Cancel where you can.
Often that strategy both works and tells you intuitively why the limit is what it is. You rarely get that from L'Hopital.
