Prove that as n -> ∞, this becomes Cos(x)?

lim(n -> ∞) (1/2)(1/k^k) (a^k + b^k)

= (1/2) lim(n -> ∞) [(a/k)^k + (b/k)^k]

= (1/2) lim(n -> ∞) ([x/(4n) + √((x/4n)^2 - 1)]^(4n) + [x/(4n) - √((x/4n)^2 - 1)]^(4n))

= (1/2) lim(y -> 0) ([y + √(y^2 - 1)]^(x/y)) + [y - √(y^2 - 1)]^(x/y))), letting y = x/(4n)

= (1/2) lim(y -> 0) (exp{x • ln(y + √(y^2 - 1))/y} + exp{x • ln(y - √(y^2 - 1))/y})

= (1/2) lim(y -> 0) (exp{x • ln(y + √(y^2 - 1)/y} + exp{-x • ln(y + √(y^2 - 1))/y})

= (1/2) (exp{x • lim(y -> 0) ln(y + √(y^2 - 1)/y} + exp{-x • lim(y -> 0) ln(y + √(y^2 - 1)/y}

= (1/2) (exp{x • lim(y -> 0) 1/√(y^2 - 1)} + exp{-x • lim(y -> 0) 1/√(y^2 - 1)}), by L'hospital's rule

= (1/2) (exp{x • 1/i} + exp{-x • 1/i})

= (1/2) (exp{-ix} + exp{ix})

= cos(x).

@Scythian thanks for spotting the mistake.

Note that ab = k^2 where k = 4n is an even integer.

So (1/2) (a^k + b^k)/k^k =

(1/2) (a^k + b^k) / (ab)^(k/2) =

(1/2) ((a/b)^(k/2) + (b/a)^(k/2))

Since n, and therefore k, is going to approach infinity, we may

as well write

a = x + i√(k² - x²) and b = x - i√(k² - x²) with |a| = |b| = k

Then we can assume that

a = k exp(i θ) and b = k exp(-i θ)

where θ = arccos(x/k) for large enough values of k.

Then (1/2) (a^k + b^k)/k^k =

(1/2) ((a/b)^(k/2) + (b/a)^(k/2)) =

(1/2) (exp(i k θ) + exp(- i k θ)) =

cos(k θ)

I was doing great up to here and then things fell apart

For large enough k let x/k=sin(α) so α→0 as k→∞

Writing x={ sin(α)/α }{ kα } and taking limits as k→∞ gives x = 1*lim(kα) = lim(kα)

From definitions of a & b, a/k=exp(i(π/2−α)) and b/k=exp(−i(π/2−α))

∴ ½( (a/k)ᵏ + (b/k)ᵏ ) = cos(k(π/2−α)) = cos(kα) since k is a multiple of 4

lim cos(kα) = cos( lim(kα) ) = cos(x)

If k=4n+1, 4n+2, 4n+3 the limit is sin(x), −cos(x),−sin(x)

That's a pretty powerful function you have there.