1^ste manier : de snelste en eenvoudigste manier
\(\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}=\lim_{x\to\,0}\:\left(1+\frac{\sin^3 x}{x^3}\right)=1+1^3=2\)
2^de manier : met de regel van de l'Hospital
\(\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\:\frac{3x^2+3\sin^2x.\cos x}{3x^2}\\=1+\lim_{x\to\,0}\:\frac{\sin^2x.\cos x}{x^2}=1+1^2.1=2\)
3^de manier : zonder de regel van De L'Hospital
\(\small\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}=\lim_{x\to\,0}\:\frac{x+\sin x}{x}\cdot\lim_{x\to\,0}\:\frac{x^2-x.\sin x+\sin^2x}{x^2}\\=(1+1)\cdot\lim_{x\to\,0}\:\left(1-\frac{\sin x}{x}+\frac{\sin^2x}{x^2}\right)\\=2\cdot(1-1+1^2)=2\cdot 1=2\)