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Attached is a graph of a function of $x$ with three positive parameters $a_1$ $a_2$ $a_3$. The dashed horizontal line is $y=a_3$. Upon visual inspection of this graph, clearly this function approaches $y=a_3$ as x goes to infinity. However, looking at the function itself, I cannot see how this is happening:

$$f(x;a_1,a_2,a_3)=(1+u)\left(\left(\frac{1+u}{u}\right)^{a_3}-1\right)$$ where $u=\left(\frac{x}{a_1}\right)^{a_2}$

It seems to me that the limit as x goes to infinity should be zero. I'd appreciate it if someone can point out why the graph of the function is approaching $y=a_3$. (The parameter values in the graph are: $a_1=0.5$ $a_2=2.1$ and $a_3=1.5$)

enter image description here

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It's an interesting take to describe this a discrepancy between the function and its graph. From a Bayesian perspective, the likelihood of a discrepancy between the function and your understanding of the function seems many orders of magnitude higher than the likelihood of a discrepancy between the function and its graph :-) – joriki Oct 31 '12 at 12:48
If you were concluding that the limit should be zero because the right hand factor tends to zero, that fails because the left hand factor is going to infinity. So, it is an indeterminate form $\infty\cdot 0$. – rschwieb Oct 31 '12 at 13:09
@joriki Obviously, but try fitting all that into the title bar. – ben Oct 31 '12 at 13:23
up vote 0 down vote accepted

You can write $\left(\frac{1+u}u\right)^{a_3}=\left(1+\frac 1u\right)^{a_3}\approx 1+\frac {a_3}u$, where the approximation is valid as $u$ (and therefore $x) \to \infty$. Then your function goes to $a_3$ as your graph does.

Added: Inside the outer parentheses you then have $1+\frac {a_3}u -1=\frac{a_3}u$. This is multiplied by the outer $(1+u)$, giving $a_3+\frac {a_3}u$. The first term is your constant.

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Thanks for the answer. I still don't understand. Doesn't $1+\frac{a_3}{u}$ go to $1+0$ as u (and x) go to infinity? And doesn't this then result in the right factor going to $0$? – ben Oct 31 '12 at 13:21
@ben: See addition – Ross Millikan Oct 31 '12 at 13:34
Wow that's pretty neat. Thanks. – ben Oct 31 '12 at 13:58

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