What Happens in the Lowest Order Approximation When \( ka \ll 1 \)?

[dustydude]

Homework Statement

I trying to figure this out, its part of a bigger question.

When $$ka \ll 1$$, what happens to,
$$\frac{1}{\left ( 2-k^{2}a^{2} \right )\textup{sin}\, ka\, - 2\, ka \textup{cos}\, ka}$$

Homework Equations

Its something to do with the lowest order approximation.

The Attempt at a Solution

I have that
$$\textup{cos}\, ka\approx 1$$
$$\textup{sin}\, ka\approx ka$$

which would make the denominator equal 0.
Is this right?

 

[Char. Limit]

Homework Statement

I trying to figure this out, its part of a bigger question.

When $$ka \ll 1$$, what happens to,
$$\frac{1}{\left ( 2-k^{2}a^{2} \right )\textup{sin}\, ka\, - 2\, ka \textup{cos}\, ka}$$

Homework Equations

Its something to do with the lowest order approximation.

The Attempt at a Solution

I have that
$$\textup{cos}\, ka\approx 1$$
$$\textup{sin}\, ka\approx ka$$

which would make the denominator equal 0.
Is this right?

Not quite "equal", since the equalities there aren't actual equalities, but just approximation. However, you can, I believe, guess that the denominator would be NEAR zero, and thus the value of the function would be very large.