Brief idea of L'Hospital's Rule. Consider two functions f(x) and g(x). If x→alimf(x)=0 and x→alimg(x)=0 (a 00 indeterminate form), and if x→alimf′(x)=p and x→alimg′(x)=q where q=0, then
limx→ag(x)f(x)=limx→ag′(x)f′(x)=qp
If instead x→alimg′(x)=0 too, then — provided x→alimf′(x)=0 as well — the same 00 situation reappears one derivative down, and x→alimg′(x)f′(x) can itself be studied by the same rule (differentiating top and bottom again).
Example 1: x→0limx2sinx. Here f(x)=sinx with limx→0f(x)=0, and g(x)=x2 with limx→0g(x)=0. Then f′(x)=cosx, so limx→0f′(x)=cos0=1=0; but g′(x)=2x, so limx→0g′(x)=2(0)=0. Since limx→0g′(x)=0 (and f′'s limit is not also 0), L'Hospital's Rule cannot be applied here. …