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I want to obtain

Exp[-α/(1 - α)] /. {α -> 1}

The output is Indeterminate. But I calculate without Mathematica and I obtain 0.

$$-\frac{1}{0} \rightarrow -\infty \implies e^{-\infty} \rightarrow 0$$

For another example, please consider

Limit[-(-1 + E^((t*α)/(-1 + α)))*x/α, α -> 1, Direction -> 1]

$x$ is real and $t > 0$.

Any suggestion?

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  • $\begingroup$ You need to take the Limit. $\endgroup$ – Marius Ladegård Meyer Jun 9 '16 at 8:28
  • $\begingroup$ @ Marius Ladegård Meyer I have problem with this limit Limit[-(((-1 + E^((t*\[Alpha])/(-1 + \[Alpha])))*x)/\[Alpha]), \[Alpha] -> 1, Direction -> 1] $\endgroup$ – user37694 Jun 9 '16 at 8:35
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    $\begingroup$ Please add that info to the question. What do you know about x and t? Are they real? Are they positive? Negative? Add that to the question also. $\endgroup$ – Marius Ladegård Meyer Jun 9 '16 at 8:47
  • $\begingroup$ @Marius Ladegård Meyer Many thanks. I edited question. $\endgroup$ – user37694 Jun 9 '16 at 8:53
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If you provide those important Assumptions to Limit, it will correctly compute that the exponential goes to zero, as you already know:

Limit[
 -(((-1 + E^((t*α)/(-1 + α)))*x)/α), α -> 1, 
 Direction -> 1, Assumptions -> t > 0 && Element[x, Reals]
]
(* x *)

This is explained in the documentation for Limit -> Examples -> Options -> Assumptions. By the way, I think one usually prefers to use Exp[x] instead of E^(x).

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  • $\begingroup$ Lukas, why would it make a difference whether one uses Exp[x], or E^x? As far as I know, they are perfectly equivalent, and one of the two is converted into the other automatically anyway. $\endgroup$ – MarcoB Jun 9 '16 at 16:59
  • $\begingroup$ @MarcoB I honestly think it is cleaner and easier to read. Might be just my opinion... But for instance in this example there are already so many parentheses that the square brackets for Exp clearly separate the exponential from the rest. You're of course right with that it does not affect computations :) $\endgroup$ – Lukas Jun 9 '16 at 17:08

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