Example — First-order substitution

limx0sin(3x)x=limx03xx=3.\lim_{x\to 0}\frac{\sin(3x)}{x} = \lim_{x\to 0}\frac{3x}{x} = 3. Here sin(3x)3x\sin(3x)\sim 3x (with =3x0\square = 3x\to 0), a legitimate substitution because sin(3x)\sin(3x) is a factor.

Example — When the second order is needed

limx0ex1xx2.\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}. The approximation ex1xe^x-1\approx x is not enough: it would give 0/00/0, because ex1xe^x-1-x is a sum in which the first-order term cancels. The second order is needed: ex1+x+x22e^x \approx 1+x+\dfrac{x^2}{2}, whence ex1xx22e^x-1-x\approx \dfrac{x^2}{2}, and the limit is 12\dfrac{1}{2}.

Comparing the two examples reveals the practical rule: substitute at first order when the function is an isolated factor; go up to second order when the first-order terms cancel out in a sum.

Topics: Limits
Concepts: Indeterminate forms · Equivalent functions
Skills: Calculating limits