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

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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?
 
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dustydude said:

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.
 
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