f(x) is differentiable for all x in [0, 1] and f(0) = 0. If a(n) = n f(1/n), show that the limit of a(n) as n approaches infinity is f ' (0)
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lim(n→∞) a(n)
= lim(n→∞) n f(1/n)
= lim(n→∞) [f(1/n) - f(0)] / (1/n), since f(0) = 0
= lim(t→0+) [f(t) - f(0)] / t, letting t = 1/x
= lim(t→0+) [f(t) - f(0)] / (t - 0)
= f '(0), since f is differentiable.
I hope this helps!
= lim(n→∞) n f(1/n)
= lim(n→∞) [f(1/n) - f(0)] / (1/n), since f(0) = 0
= lim(t→0+) [f(t) - f(0)] / t, letting t = 1/x
= lim(t→0+) [f(t) - f(0)] / (t - 0)
= f '(0), since f is differentiable.
I hope this helps!