This may be a stupid question: The function $ \frac{x}{1-e^x} $, which positive and negative limits around $0$ are $-1$, is defined or not for $x = 0$?
Domain of $x/(1-e^x)$
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calculus
limits
functions
3 Answers
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No, it's not defined at $x = 0$. However, the function $$ g(x) = \cases{\frac x{1-e^x} & if $x \neq 0$\\-1 & if $x = 0$} $$ is mathematically indistinguishable from your function for all non-zero inputs, but it's defined and continuous at $x = 0$ as well.
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2I think you want $g(0) = -1$. :) – 2017-01-14
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0Thank you for the complete answer! – 2017-01-14
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it's not defined at zero as you would be dividing by zero
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Definitely not defined for $x = 0$, since $e^0 = 1$, so $1 - e^0 = 0$, and division by $0$ is undefined.
On the other hand, there IS a contininuous function $g$ that equals your function for every nonzero $x$, and for which $g(0) = -1$.