# Simplifying an expression using limits

1. Jan 16, 2016

### whatisreality

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

$T = \frac{4E(V-E)}{4E(V-E)+V^2\sinh^2 (ka)}$

Simplify the expression for T when a is large but not infinite, and again for the case when a tends to zero and the potential tends to infinity, such that $d = Va$ is a real constant. k is a constant, as is E (which is assumed to be less than V).

2. Relevant equations

3. The attempt at a solution
To try and simplify it for the first case, I used the exponential definition of sinh, squared it and got that
$\sinh^2(x) = \frac{1}{4}(e^{2x}+e^{-2x}-2)$

For large x the $e^{-2x}$ term would get small very quickly, especially compared to the other exponential term which would increase rapidly. So this becomes

$\sinh^2(x) = \frac{1}{4}(e^{2x}-2)$

I don't know if that's what the question meant. It doesn't seem massively simpler. Is there something else that can be done to simplify the expression?

And the second part, I don't know where to start! There should be some sort of constant $d = Va$, but I thought $sinh^2(0)=0$, which would mean that $a$ disappears. As far as I can see that expression simplifies to 1 for the second set of limits! Why is that wrong?

2. Jan 16, 2016

### Fightfish

What is the dominating term in the denominator?
The limit is such that a tends to zero while V tends to infinity at the same time, so the limit of the term $V^2 \sinh^{2}(ka)$ cannot be done simply by setting a = 0 as you've done. A good starting point is to Taylor expand $\sinh^{2}(ka)$, and see where that gets you to, before taking the limits.

3. Jan 16, 2016

### Ray Vickson

Even better in the large-$a$ case would be to write $\sinh^2(x) \sim e^{2x}/4$, because $e^{2x}$ is overwhelmingly large compared to the 2. Also, if $E(E-V)$ remains constant, it is overwhelmingly small compared with the second term in the denominator, and that leads to even more simplification.

4. Jan 16, 2016

### vela

Staff Emeritus
It would help to note that
$$k^2 = \frac{2m(V-E)}{\hbar^2}.$$ Since $ka$ depends on both $a$ and $V$, you need to first figure out how it behaves in the limit as $a \to 0$ and $V \to \infty$ before writing down an approximation for $\sinh ka$. Start by writing $V = d/a$ and getting everything in terms of $a$.

5. Jan 16, 2016

### whatisreality

Oh, I completely forgot about that! Thanks, I'll rewrite in terms of $a$ now.

6. Jan 20, 2016

### whatisreality

In terms of a, $ka = \frac{\sqrt{2mad - 2ma^2E}}{\hbar}$.

That still tends to zero, given d is a real constant. So $sinh^2(ka)$ tends to zero, since its taylor expansion is $x^2 + \frac{x^4}{3} + \frac{2x^6}{45}$.

Taking the $V^2$ term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the $sinh^2$ term tends to zero!

Last edited by a moderator: Jan 20, 2016
7. Jan 20, 2016

### whatisreality

Sorry, misplaced # in the post above.

In terms of a, $ka = \frac{\sqrt{2mad - 2ma^2E}}{\hbar}$.

That still tends to zero, given d is a real constant. So $sinh^2(ka)$ tends to zero, since its taylor expansion is $x^2 + \frac{x^4}{3} + \frac{2x^6}{45}$.

Taking the $V^2$ term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the $sinh^2$ term tends to zero!

8. Jan 20, 2016

### whatisreality

Is getting everything in terms of a equivalent to getting it in terms of V, or do both cases have to be considered?

9. Jan 20, 2016

### Fightfish

Why not try multiplying $V^{2}$ to your Taylor series expansion and see what you get? There'll be some simplifications that allow you to take the limit properly.
[Hint: use Va = d]