I am given $\omega\left ( k \right )=\left ( \frac{C+H}{m}\pm \frac{1}{m}\sqrt{C^{2}+H^{2}+2 CH Cos\left ( k a \right )} \right )^{1/2}$.
It is mentioned that this reduces to
$\omega\left ( k \right )=\sqrt{\frac{CH}{2m\left ( C+H \right )}}a\left | k \right |$ when $\lambda \rightarrow \infty$ for $\lambda = \frac{2 \pi}{k}$.
After 5 hours of frustrating attempt, I am unable to progress further.
Attempt:
Note that in the event that $\lambda \rightarrow \infty, k\rightarrow 0$.
Thence, it can be expected that $ka \approx 0$ and so $Cos\left ( ka \right )\approx 1$.
It is useful to recall that for small angle $\left ( ka \right ), Cos\left ( ka \right )\approx 1-\frac{\left ( ka \right )^{2}}{2}$.
Expanding this we obtain at
$\left ( \frac{C+H}{m}-\frac{1}{m}\sqrt{C^{2}+H^{2}+2CH} \right ) =0$
which isn't what I am looking for.
Any help is appreciated.
Hint
As you wrote, you need to use $$\cos(ka)=1-\frac{a^2 k^2}{2}+O\left(a^4\right)$$ which makes $$C^{2}+H^{2}+2 CH \cos\left ( k a \right )=(C+H)^2-CH a^2k^2+O\left(a^4\right)$$ $$\sqrt{C^{2}+H^{2}+2 CH \cos\left ( k a \right )}\sim(C+H)\sqrt{1-\frac{CHa^2k ^2}{(C+H)^2}}\sim(C+H)\left(1-\frac{CHa^2k ^2}{2(C+H)^2} \right)$$
Just continue.