Is it possible to make a function of the form $A_1 e^{\frac{A_2}{\sqrt{x - A_3}}}$ converge to $B_1 e^{-B_2x}$ by using the correct constants?

Question

There is a function $$ f_1 = B_1 e^{-B_2x} $$ where $B_1$ and $B_2$ are constants, and $x$ is a variable.

Is it possible to make a function of the form $$f_2 = A_1 e^{\frac{A_2}{\sqrt{x - A_3}}}$$ where $A_1$, $A_2$, and $A_3$ are constants converge to $ B_1 e^{-B_2x} $?

In other words say I plot $f_1$. Can I plot $f_2$ with appropriate constants so that $f_1$ and $f_2$ are almost identical?

I think it may be possible, but I am not sure. To elaborate, we can set $A_3 = - \infty$ to push the vertical asymptote of $f_2$ to $- \infty$ as $f_1$ does not have a vertical asymptote. And I think it should be possible to find constants $A_1$ and $A_2$ such that $f_1$ and $f_2$ look almost identical.

Answer 1

The function $f_1$ satisfies $$ \dfrac{f_1'(x)}{f_1(x)} = - B_2 $$ while for $f_2$ $$ \dfrac{f_2'(x)}{f_2(x)} = - \frac{A_2}{(x-A_3)^{3/2}} $$ which is not constant. However, for plotting on a fixed finite interval we can take $-A_3$ large compared to $x$ and adjust $A_2$ to make $f_2'/f_2$ close to $-B_2$ on that interval. Then with an appropriate $A_1$, the functions will be close on this interval.

For example, take $B_1 = B_2 = 1$ and let the interval be $[-1,1]$. If I take $A_3 = -400$, $A_2 = 16000$, I will have $f_2'/f_2$ between approximately $-0.99962$ and $-1.00038$ on the interval, and with $A_1 = e^{-800}$ the functions are equal at $x=0$. The plots of $f_1$ and $f_2$ on the interval $[-1,1]$ will be practically indistinguishable: the maximum absolute value of the difference between them is at most about $0.005$.