Approximating $\lim_{x\to\infty} (1-c^{-1})^x$

Question

I have encountered the following approximation:

given that $c\le x$, $\lim_{x\to\infty} (1-c^{-1})^x \approx (1-x^{-1})^{x\frac{c}{x}} \approx e^{-\frac{c}{x}}$.

Can anyone explain the derivation of this approximation.

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Edit: Here are some context:

Answer 1

If you know that $\ln(1+x) \approx x$ for $x$ close to $0$ you could reason that $$(1-c^{-1})^x = e^{x \ln (1 - \frac 1c)} \approx e^{-\frac cx}$$ provided that $1/c$ is close to $0$, that is, $c$ large.