Extreme sensitivity to the input scale of a Continuous Fourier Transform approximation from Discrete Fourier Transform

Is there a way to 'bypass' the condition $$\Delta k=\frac{1}{x_{max}-x_{min}}$$, which requires the range of $$x$$ to be very large when approximating the CFT encountered in Michealson Interometery as a DFT?

To contextualise my question:

Suppose the the continuous Fourier Transform (CFT) is defined as $$\hat{f}(k)=\frac{1}{2}\int_{-\infty}^{\infty} f(x) e^{-i2\pi kx}dx$$

Suppose that $$f \to 0$$ for sufficiently large and small values of $$x$$, then $$\hat{f}(k)=\frac{1}{2}\int_{-\infty}^{\infty} f(x) e^{-i2\pi kx}dx$$ $$\hat{f}(k_m)\approx \frac{1}{2}\sum_{n=0}^{N-1}f(x_n)e^{-i2\pi k_mx_n}\Delta x$$ $$=\frac{1}{2}\sum_{n=0}^{N-1}f(x_0+n\Delta x)e^{-2i(m\Delta k)(x_0+n \Delta x)}\Delta x$$ $$=\frac{1}{2}e^{-2i m\Delta k x_0 }\sum _{n=0}^{N-1}f(x_0+n\Delta x)e^{-i2mn \Delta x \Delta k }$$ where $$x_n=x_0+n\Delta x$$, $$k_m=m \Delta k$$ (so $$k_0=0$$)

The Discrete Fourier Transform is defined as follows $$F_m=\sum^{N-1}_{n=0}f_ne^{-i2mn/N}$$ We conclude that for the last sum appearing in the CFT to approximate to the DFT, $$\Delta x \Delta k=1/N$$ must hold. $$\hat{f}(k_m)\approx \frac{1}2{e^{-2im\Delta kx_0}F_m} \tag{1}$$

since $$\Delta x=\frac{x_{max}-x_{min}}{N-1}$$, $$\Delta k=\frac{1}{N \Delta x} \approx \frac{1}{x_{max}-x_{min}}$$ for N sufficiently large.

This means that if I use the approximation (1) in the nano-metre scale, it will fail since $$\Delta k$$ (separation between $$k_{i+1}$$ and $$k_{i}$$)will be very large. So (1) must be used in the range, say $$-100, a range like $$-10^-{9} (visible wavelength scale) will cause the approximation to break down.