# Computing a ratio of exponential functions without overflow issues

I'm interested in computing pointwise values of the function $$u(x) = \sinh(k-kx)/\sinh(k)$$ for $$x \in (0,1)$$, where $$k = 10^{4}$$. A direct computation of course results in overflow issues due to the $$\exp(k)$$ factor. However, $$u(x)$$ only takes on values between 0 and 1, so I'm wondering if there is a clever way that this quantity can be computed. We can also write $$u$$ as $$u(x) = \frac{e^{k-kx}}{e^{k}-e^{-k}} - \frac{e^{kx-k}}{e^{k}-e^{-k}}.$$

I thought to try taking logarithms first and seeing if anything simplifies, but I wasn't able to make progress. Does anyone have suggestions for how I can get numerical pointwise values of this function? I'm aware that the function is essentially 1 at $$x=0$$ and 0 elsewhere, but I'm curious about the original problem.

My idea is the approximation $$\exp(k)-\exp(-k)\approx \exp(k)$$ for large positive values of $$k$$. But instead, we can just divide the numerator and denominator by $$\exp(k)$$ and get an exact expression $$\frac{\exp(k-kx)-\exp(kx-k)}{\exp(k)-\exp(-k)} = \frac{\exp(-kx)-\exp(k(x-2))}{1-\exp(-2k)}$$ which should be okay to evaluate without overflow. (Note this is exactly the same technique used to evaluate the log-sum-exp function without overflow.)
Under assumptions you give for $$x$$ and I assume $$k$$ to be positive as well, you can calculate an asymptotic expansion for this function for $$k\to\infty$$: $$u(x) \sim e^{-kx} - e^{-k(2-x)} + e^{-k(2+x)} -e^{-k(4-x)} +e^{-k(4+x)} + \ldots$$ from which you can see that if $$k$$ is large enough (and $$10^{4}$$ certainly is), only the first term is significant.