Let c be a real number. Let f be defined for x in (c,infinity) and f(x) > 0 for all x in (c,infinity).?

Question

Show that lim as x->c of f = infinity if and only if lim as x approaches c of 1/f = 0.

Answer 1

(=>) Suppose lim (x->c) f(x) = ∞ and let ε > 0. Then let N = 1/ε. Since lim (x->c) f(x) = ∞ there exists δ > 0 such that for c < x < c+δ, f(x) > N > 0. So for c < x < c+δ, 1/f(x) < 1/N = ε. So lim (x->c) 1/f(x) = 0.

(<=) Suppose lim (x->c) 1/f(x) = 0 and let N > 0. Then let ε = 1/N. Since lim (x->c) 1/f(x) = 0 there exists δ > 0 such that for c < x < c+δ, |1/f(x)| < ε. Since f(x) > 0 for all x > c, we have 0 < 1/f(x) < ε. So f(x) > 1/ε = N. So lim (x->c) f(x) = ∞.