Complications Limits of Powers

Question

Sometimes L'Hôpital's rule does not lead to an answer in a finite number of steps unless a transformation of variables is applied.

How can I find the limit of the function $f(t)$:

$$\lim_{t\to \infty}f(t)=\lim_{t\to \infty}\frac{a e^{-at}+b e^{-bt}-(a+b)e^{-(a+b)t}}{e^{-at}+ e^{-bt}-e^{-(a+b)t}},$$ where $a\geq0$ and $b\geq 0$.

My attempts $$\lim_{t\to \infty}f(t)=\lim_{t\to \infty}\frac{a e^{at}+b e^{bt}-(a+b)}{e^{at}+ e^{bt}-1}$$

if $b=a$.

Either $$\lim_{t\to \infty}f(t)=\lim_{t\to \infty}\frac{a-\frac{2a}{e^{at}+e^{bt}}}{1-\frac{1}{e^{at}+e^{bt}}}=a,$$ or $$\lim_{t\to \infty}f(t)=\lim_{t\to \infty}\frac{b-\frac{2b}{e^{at}+e^{bt}}}{1-\frac{1}{e^{at}+e^{bt}}}=b.$$

Is this route logical?

Is there another way to prove?

thanks for the help.

Answer 1

If $a>b\geq0$ we obtain: $$\lim\limits_{t\rightarrow+\infty}\frac{a e^{-at}+b e^{-bt}-(a+b)e^{-(a+b)t}}{e^{-at}+ e^{-bt}-e^{-(a+b)t}}=\lim\limits_{t\rightarrow+\infty}\frac{a e^{(b-a)t}+b-(a+b)e^{-at}}{e^{(b-a)t}+ 1-e^{-at}}=b.$$ If $a=b$ we obtain $$\lim\limits_{t\rightarrow+\infty}\frac{a e^{-at}+b e^{-bt}-(a+b)e^{-(a+b)t}}{e^{-at}+ e^{-bt}-e^{-(a+b)t}}=\lim\limits_{t\rightarrow+\infty}\frac{2a e^{-at}-2ae^{-2at}}{2e^{-at}-e^{-2at}}=\lim\limits_{t\rightarrow+\infty}\frac{2a -2ae^{-at}}{2-e^{-at}}=a.$$ If $b>a\geq0$ we obtain: $$\lim\limits_{t\rightarrow+\infty}\frac{a e^{-at}+b e^{-bt}-(a+b)e^{-(a+b)t}}{e^{-at}+ e^{-bt}-e^{-(a+b)t}}=\lim\limits_{t\rightarrow+\infty}\frac{a +b e^{(a-b)t}-(a+b)e^{-bt}}{1+ e^{(a-b)t}-e^{-bt}}=a.$$ The similar things we'll get for $t\rightarrow-\infty$.

Answer 2

If $a,b>0$, we factor out $e^{-\min(a,b)t}$ from both denominator and numerator, then we find that $\lim_{t\to\infty} f(t) = \min(a,b)$. If one of them vanishes, w.l.o.g. let $a=0$, then $\lim_{t\to\infty} f(t) = 0$, such that $\lim_{t\to\infty} f(t) = \min(a,b)$ for all $a,b\geq 0$.