We can also prove (cf. § 75) that limn(1x1/n)=limn(x1/n1)=logx.

For n(x1/n1)n(1x1/n)=n(x1/n1)(1x1/n), which tends to zero as n, since n(x1/n1) tends to a limit (§ 75) and x1/n to 1 (Ex. XXVII. 10). The result now follows from the inequalities (3) of § 208.

Example LXXXVI
1. Prove, by taking y=1 and n=6 in the inequalities (4) of § 208, that 2.5<e<2.9.

2. Prove that if t>1 then (t1/nt1/n)/(tt1)<1/n, and so that if x>1 then 1xdtt1(1/n)1xdtt1+(1/n)<1n1x(t1t)dtt=1n(x+1x2). Hence deduce the results of § 209.

3. If ξn is a function of n such that nξnl as n, then (1+ξn)nel. [Writing nlog(1+ξn) in the form l(nξnl)log(1+ξn)ξn, and using Ex. LXXXII. 4, we see that nlog(1+ξn)l.]

4. If nξn, then (1+ξn)n; and if 1+ξn>0 and nξn, then (1+ξn)n0.

5. Deduce from (1) of § 208 the theorem that ey tends to infinity more rapidly than any power of y.


208. The representation of ex as a limit Main Page 210. Common logarithms