Hi,
I'm not sure how to find out the answer to this qustion so any help would be great. I have a hunch that K may be -1 and l may be +/-4.
Thanks, DLL.
The denominator is going to 0 so, in order to have any finite limit, the numerator must also go to 0. Since cos(x) goes to 1 as x goes to 0, yes, k must be -1.
You then have $\displaystyle \frac{-1+ cos(lx)}{x^2}$. I would suggest using L'Hopital's rule: differentiating numerator and denominator separately, we get $\displaystyle \frac{-l sin(lx)}{2x}= -\frac{l^2}{2}\frac{sin(lx)}{lx}$ where I have multiplied numerator and denominator by l to get "sin(z)/z" as z goes to 0.
Maclaurin series: $\displaystyle \cos lx=1-\frac{(lx)^2}{2}+O(x^4)$
Put that into the limit and you get $\displaystyle \lim_{x \to 0}\frac{k+1-\frac{(lx)^2}{2}+O(x^4)}{x^2}=\frac{k+1}{x^2}-\frac{\frac{(lx)^2}{2}+O(x^4)}{x^2}=-4$
We can see that $\displaystyle k=-1$ since otherwise the limit wouldn't be finite.
With that out of the way, we have $\displaystyle \lim_{x \to 0}-\frac{\frac{(lx)^2}{2}+O(x^4)}{x^2}=\lim_{x \to 0}\left(-\frac{l^2}{2}+O(x^2)\right)=-4 \Longleftrightarrow l^2=8$ since $\displaystyle \lim_{x \to 0} O(x^2)=0$
If you're going to use l'Hopital's rule once, you may as well use it twice.
Using it the second time gives $\displaystyle \lim_{x \rightarrow 0} \frac{-l^2 \cos (lx)}{2}$. So $\displaystyle \frac{-l^2}{2} = -4$ ....
By the way, you're expected to know that $\displaystyle \lim_{x \rightarrow 0} \frac{\sin z}{z} = 1$ ....