B. The calculation of certain limits. Suppose that f(x) and ϕ(x) are two functions of x whose derivatives f(x) and ϕ(x) are continuous for x=ξ and that f(ξ) and ϕ(ξ) are both equal to zero. Then the function ψ(x)=f(x)/ϕ(x) is not defined when x=ξ. But of course it may well tend to a limit as xξ.

Now f(x)=f(x)f(ξ)=(xξ)f(x1), where x1 lies between ξ and x; and similarly ϕ(x)=(xξ)ϕ(x2), where x2 also lies between ξ and x. Thus ψ(x)=f(x1)/ϕ(x2). We must now distinguish four cases.

(1) If neither f(ξ) nor ϕ(ξ) is zero, then f(x)/ϕ(x)f(ξ)/ϕ(ξ).

(2) If f(ξ)=0, ϕ(ξ)0, then f(x)/ϕ(x)0.

(3) If f(ξ)0, ϕ(ξ)=0, then f(x)/ϕ(x) becomes numerically very large as xξ: but whether f(x)/ϕ(x) tends to  or , or is sometimes large and positive and sometimes large and negative, we cannot say, without further information as to the way in which ϕ(x)0 as xξ.

(4) If f(ξ)=0, ϕ(ξ)=0, then we can as yet say nothing about the behaviour of f(x)/ϕ(x) as x0.

But in either of the last two cases it may happen that f(x) and ϕ(x) have continuous second derivatives. And then f(x)=f(x)f(ξ)(xξ)f(ξ)=12(xξ)2f(x1),ϕ(x)=ϕ(x)ϕ(ξ)(xξ)ϕ(ξ)=12(xξ)2ϕ(x2), where again x1 and x2 lie between ξ and x; so that ψ(x)=f(x1)/ϕ(x2). We can now distinguish a variety of cases similar to those considered above. In particular, if neither second derivative vanishes for x=ξ, we have f(x)/ϕ(x)f(ξ)/ϕ(ξ).

It is obvious that this argument can be repeated indefinitely, and we obtain the following theorem:

suppose that f(x) and ϕ(x) and their derivatives, so far as may be wanted, are continuous for x=ξ. Suppose further that f(p)(x) and ϕ(q)(x) are the first derivatives of f(x) and ϕ(x) which do not vanish when x=ξ. Then

(1) if p=q, f(x)/ϕ(x)f(p)(ξ)/ϕ(p)(ξ);

(2) if p>q, f(x)/ϕ(x)0;

(3) if p<q, and qp is even, either f(x)/ϕ(x)+ or f(x)/ϕ(x), the sign being the same as that of f(p)(ξ)/ϕ(q)(ξ);

(4) if p<q and qp is odd, either f(x)/ϕ(x)+ or f(x)/ϕ(x), as xξ+0, the sign being the same as that of f(p)(ξ)/ϕ(q)(ξ), while if xξ0 the sign must be reversed.

This theorem is in fact an immediate corollary from the equations f(x)=(xξ)pp!f(p)(x1),ϕ(x)=(xξ)qq!ϕ(q)(x2).

Example LVIII

1. Find the limit of {x(n+1)xn+1+nxn+2}/(1x)2, as x1. [Here the functions and their first derivatives vanish for x=1, and f(1)=n(n+1), ϕ(1)=2.]

2. Find the limits as x0 of (tanxx)/(xsinx),(tannxntanx)/(nsinxsinnx).

3. Find the limit of x{x2+a2x} as x. [Put x=1/y.]

4. Prove that limxn(xn)cscxπ=(1)nπ,limxn1xn{cscxπ(1)n(xn)π}=(1)nπ6, n being any integer; and evaluate the corresponding limits involving cotxπ.

5. Find the limits as x0 of 1x3(cscx1xx6),1x3(cotx1x+x3).

6. (sinxarcsinxx2)/x6118, (tanxarctanxx2)/x629, as x0.


149. Applications of Taylor’s Theorem to maxima and minima Main Page 151. The contact of plane curves