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I have a question regarding an exponential function when time goes towards infinity. The equation I have is the following:

$R^3=\frac{1-b e^{\frac{(b-1)t}{a}}}{1-b}$ $\\ \\$ $equation 1$

where $e$ represents the exponential function. $a$, $b$ and $c$ are just constants.

Then the book tells me that when time goes to infinity, the above expression gives me: $R=(1-b)^{-\frac{1}{3}}$.

I find this very strange, since the numerator of equation 1 should not approach 1 when time goes to infinity. Does anyone understand why this is the case here?

1 Answers 1

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The behavior of the given function as time 't' tends to infinity entirely depends upon the sign of the constant $\frac{(b-1)}{a}$. If $\frac{(b-1)}{a}$ is positive, then it is impossible to get a finite answer, as in your case, as and when time 't' tends to infinity.

However, if $\frac{(b-1)}{a}$ is somehow negative, then it is not very difficult to see how you would arrive at the result given in your textbook.

I, personally, believe it is an error of a minus sign which is rather common when not paying enough attention.

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    It was not stated if b-1<0 or not, but I guess I have to assume this, and as you mentioned it then works out fine. Thank you! – 2017-02-03