Understanding Limit: N=1 +/- sqrt(Ae^(2rt))/sqrt(1-Ae^(2rt))

[binbagsss]

Homework Statement

[/B]
Trying to understand this limit:

where $r>0$

Homework Equations

[/B]
I think it's best to proceed by writing this as:

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

The Attempt at a Solution

[/B]
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'hospital's rule being much use here due to the square root and exponential terms.

Many thanks in advance.

 

[Ray Vickson]

Homework Statement

[/B]
Trying to understand this limit:
View attachment 217254
where $r>0$

Homework Equations

[/B]
I think it's best to proceed by writing this as:

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

The Attempt at a Solution

[/B]
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'hospital's rule being much use here due to the square root and exponential terms.

Many thanks in advance.

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

 

[binbagsss]

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

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

 

[Ray Vickson]

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

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

 

[binbagsss]

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

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