For what value of k is the lim x>infinity for sinh(kx)/cosh(5x) finite

sinh(kx)/cosh(5x)

I solved it down to e^(k-5)x
Finite value defined as k< or = to 0
so solving for it gives me a value of 5
The interval the way I solved it should look like (-infinity,5) however I'm getting it wrong.

Did I misstep somewhere?

Since sinh(-kx) = -sinh(kx), assume for now that k is non-negative.

Moreover, sinh 0 = 0, this limit is finite if k = 0.
So, assume that k > 0.

Note that
sinh(kx)/cosh(5x) = (e^(kx) - e^(-kx)) / (e^(5x) + e^(-5x))

Since lim(x→∞) e^(-nx) = 0 if n > 0,
lim(x→∞) sinh(kx)/cosh(5x) is finite
<==> lim(x→∞) e^(kx) / e^(5x) is finite
<==> k ≤ 5.

This, with the symmetry comment at the beginning, implies that -5 ≤ k ≤ 5.

I hope this helps!