1^ste manier : zonder de regel van de L'Hospital
\(\displaystyle\lim_{x\to 0}\frac{1\!-\cos x\,+\,x^2}{1\:-\:\cos x}=\lim_{x\to 0}\frac{1\!-\!\cos x}{1\!-\!\cos x}+\displaystyle\lim_{x\to 0}\frac{x^2}{1-\cos x}=1+\lim_{x\to 0}\frac{x^2(1+\cos x)}{(1-\cos x)(1+\cos x)}\\=1+\displaystyle\lim_{x\to 0}\frac{x^2(1+\cos x)}{1-\cos^2x}=1+\displaystyle\lim_{x\to 0}\frac{1+\cos x}{\frac{\sin^2x}{x^2}}=1+\displaystyle\lim_{x\to 0}\frac{1+\cos x}{\frac{\sin x}{x}.\frac{\sin x}{x}}=1+\frac{1+\cos 0}{1.1}\\=1+2=3\)
2^de manier : met de regel van de L'Hospital
\(\displaystyle\lim_{x\to 0}\frac{1\!-\cos x\,+\,x^2}{1\:-\:\cos x}=\lim_{x\to 0}\frac{1\!-\!\cos x}{1\!-\!\cos x}+\lim_{x\to 0}\frac{x^2}{1-\cos x}\left(=1+\frac00\right)\overset{H}{=}1+\lim_{x\to 0}\frac{2x}{0+\sin x}\\=1+2.\lim_{x\to 0}\frac{x}{\sin x}=1+2.1=3\)
3^de manier : met de regel van de L'Hospital
\(\displaystyle\lim_{x\to 0}\frac{1\!-\cos x\,+\,x^2}{1\:-\:\cos x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to 0}\frac{\sin x+2x}{\sin x}=1+2.\lim_{x\to 0}\frac{x}{\sin x}=1+2.1=3\)