What are the asymptote(s) and hole(s), if any, of $f(x) =(1-e^-x)/x$ ?

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Answer 1 · 2016-10-02T23:06:58

The only asymptote is $x$ =0

Of course, $x$ cannot be 0, otherwise $f(x)$ remains undefined. And that's where the 'hole' in the graph is.

What are the asymptote(s) and hole(s), if any, of f(x) =(1-e^-x)/x f(x) =(1-e^-x)/x ?