Homework Help: Quick Limits Question

1. Dec 22, 2017

binbagsss

1. The problem statement, all variables and given/known data

Trying to understand this limit:

where $r>0$
2. Relevant equations

I think it's best to proceed by writing this as:

$N=1 \pm \frac{\sqrt{Ae^{2rt}}}{\sqrt{1-Ae^{2rt}}}$

3. The attempt at a solution

since $r>0$ the exponential term $\to$ $\infty$ and then since $A<0$ I get two results for $lim_{t \to \infty} Ae^{2rt}$ depending on $|A|$.

a) If $|A| < 1$ it goes to zero.
if b) $|A| \geq 1$ it goes to $-\infty$

and where the magnitude of A is not specified in the question.

If it was however for case a) the limit is of an determinate form: $1 \pm \frac{0}{1} = 1$

however for b) i get $1 \pm \frac{\sqrt{-\infty}}{\sqrt{1+\infty}}$ , and I can't see L'Hopitals rule being much use here due to the square root and exponential terms.

2. Dec 22, 2017

Ray Vickson

$$f(t) = \frac{\sqrt{A e^{2rt}}}{\sqrt{1-Ae^{2rt}}} = \frac{e^{-rt}}{e^{-rt}} \frac{\sqrt{A e^{2rt}}}{\sqrt{1-Ae^{2rt}}} = \frac{\sqrt{A}}{\sqrt{e^{-2rt} - A} } \to \frac{\sqrt{A}}{\sqrt{-A}}$$

3. Dec 22, 2017

binbagsss

but A<0, so the numerator is not real?

4. Dec 22, 2017

Ray Vickson

Right, but no less real than the original form $\sqrt{A e^{2rt}}$.

Last edited: Dec 22, 2017
5. Dec 22, 2017

binbagsss

so then do we not get $1 \pm i$ rather than $1 \pm 1$ as in the solution above?