Evaluating $\lim_{x\to 0}\frac{x\sin x-2+2\cos x}{x\ln(1+x)-x^2}$ using L'Hôpital Considering this limit, assigned to my high school students, 
$$\lim_{x\to 0}\frac{x\sin x-2+2\cos x}{x\ln \left(1+x\right)-x^2}=\left(\frac00\right)=\lim_{x\to 0}\frac{\frac{d}{dx}\left(x\sin \left(x\right)-2+2\cos \left(x\right)\right)}{\frac{d}{dx}\left(x\ln \left(1+x\right)-x^2\right)} \tag 1$$
After some steps, using L'Hôpital, I find:
$$\lim_{x\to 0}\frac{\left(x\cos \left(x\right)-\sin \left(x\right)\right)\left(1+x\right)}{-2x^2-x+x\ln \left(x+1\right)+\ln \left(x+1\right)}=\left(\frac00\right)$$
Should I continue to apply L'Hôpital? :-(

A quick estimate allows you to see that Taylor would yield even degree terms at the numerator (the function is even), but the constant and quadratic ones cancel each other. At the denominator, $x^2-x^2$ and cubic terms.
Hence the limit will be zero, but you will need three successive applications of L'Hospital to establish this.
$$x\sin x-2+2\cos x\to x\cos x-\sin x\to-x\sin x\to-\sin x-x\cos x\to0$$
vs.
$$x\log(1+x)-x^2\to\log(x+1)+\frac x{x+1}-2x\to\frac1{x+1}+\frac1{(x+1)^2}-2
\\\to-\frac1{(x+1)^2}-\frac2{(x+1)^3}\not\to0.$$
