the limit lemma
If the function f is differentiable at 0 and k ≠ 0, then nf(k/n) -> kf'(0) as n -> ∞
proof
From the definition of the derivative, f(x)/x -> f'(0) as x -> 0.
Provided k ≠ 0, kx -> 0 as x -> 0, so f(kx)/kx -> f'(0) as x -> 0.
Multiplying through by k, f(kx)/x -> kf'(0) as x -> 0.
Now let x = 1/n, so as n -> ∞, x -> 0, and so nf(k/n) -> kf'(0).
We shall need the result when f(x) is sin(x), sinh(x), tan(x) or tanh(x).
In each case, f'(0) = 1, so f(k/n) -> k as n -> ∞.
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