Calculus Derivative—Finding unknown constants 
Find all values of $k$ and $l$ such that
  $$\lim_{x \to 0} \frac{k+\cos(lx)}{x^2}=-4.$$

Any help on how to do this would be greatly appreciated.

Note that
$$\lim \frac{1-\cos x}{x^2}=\frac{1}{2}$$
And that $$\frac{k+\cos(lx)}{x^2}=\frac{k+1}{x^2}-\frac{1-\cos(lx)}{x^2}$$
Thus $$k+1=0$$and $$\frac{1}{2}l^2=4$$
