# Finding $\lim_{E \to U} \frac{1}{4}\left[\frac {U^2}{E(E-U)}\right]\sin^2 k'L$ where $k' = \left[2m(E-U)/\hbar\right]^{1/2}$ and $L$ is a constant

I am trying to find the limit of this equation:

$$\lim_{E \to U} \frac{1}{4}\left[\frac {U^2}{E(E-U)}\right]\sin^2 k'L$$

Isn't there no limit, because $(E-U)$ would be 0, and the denominator would become 0, which means fraction is undefined?

Edit:

$k' = \left[2m(E-U)/\hbar\right]^{1/2}$ and $L$ is some constant

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That would depend on what $k'$ and $L$ might be. – Gerry Myerson Aug 6 '12 at 11:55
I forgot to write that $k'$ is dependent on $E$... – Asalph Tooks Aug 6 '12 at 12:00

## 2 Answers

If you are familiar with $\lim_{x\to0}{\sin x\over x}=1$ then you should be able to manipulate your problem to where you can make use of this limit.

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Well you're right, but you're right for the wrong reasons. There are plenty of instances where plugging in a number to evaluate a limit gives you something undefined but the actual limit exists. For example if you take the limit as x goes to 2 of $\ \frac{x^2 - 5x + 6}{x-2}$ then you will get 0 in the denominator when you plug in 2 for x, yet you will see that this quotient approaches -1 as x goes to 2.

For your function, assuming E, U are real numbers, your trigonometric function can be regarded as a constant, and as E approaches U, the function will grow without bound from one direction (tend to positive infinity) and diminish without bound from another (tend to negative infinity). The limit cannot exist because (i) the limit from the right and the left must agree for the limit to exist (ii) the limit must be a real number (not infinity or negative infinity).

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His trigonometric function is not a constant because $k'$ depends on $E$ (but this is obvious only after his edit). – celtschk Aug 6 '12 at 12:17