Find the value of K. Use of l Hospital's rule and expansion is not allowed. Let $f(x) =\log_{cos3x} (\cos2ix)$ if $x \ne 0$ and $f(0) = k$ where ($i$= iota) is continuous at $x = 0$, then find the value of $K$. Use of l Hospital's rule and expansion is not allowed.
 A: (Preamble: Apart from addition and multiplication the only genuine two-ary function allowed in print is the general power function 
$$(x,y)\mapsto x^y:=\exp(y\, \log x)\qquad(x>0, y\in{\mathbb C})\ .\tag{1}$$
The function $(x,y)\mapsto \log_x y$ where both $x$ and $y$ are variables is not of this kind. When such a thing should really come up in practice one would have to rewrite it in such a way that only functions of one variable and the "allowed" operations appear. It is true that we sometimes write $\log_a y$. But the  $a>1$ here is not a variable; it is  a constant scaling parameter chosen in a most  suitable way for the problem at hand.)
So after a lot of mind bending I'm interpreting your problem as follows: For $0<|x|\ll1$ one has $0<\cos (3x)<1$. On the other hand
$$\cos(2ix)={e^{-2x}+e^{2x}\over2}=\cosh (2x)\qquad(x\in{\mathbb R})\ .$$
Therefore the equation
$$\left(\cos(3x)\right)^y=\cos(2ix)=\cosh(2x)$$
has a real solution $y$ for such $x$; and according to $(1)$ this $y$ is given by
$$y={\log\bigl(\cosh(2x)\bigr)\over \log\bigl(\cos(3x)\bigr)}=: f(x)\qquad(0<|x|\ll1)\ .\tag{2}$$
The right hand side of $(2)$ is the correct expression of the function $f$ your teacher had in mind. 
Now it comes to computing $\lim_{x\to0} f(x)$. Since both de l'Hôpital's rule and expansion are forbidden I have to stop here. It is unclear to me how this limit can be computed without using the analytical properties of the involved functions somehow.
