0
$\begingroup$

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

  • 0
    @joriki Obviously, but try fitting all that into the title bar.2012-10-31

1 Answers 1

0

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.

  • 0
    Wow that's pretty neat. Thanks.2012-10-31