Jovany Clayton

2022-07-12

Find the value of K. Use of l Hospital's rule and expansion is not allowed.
Let $f\left(x\right)={\mathrm{log}}_{cos3x}\left(\mathrm{cos}2ix\right)$ if $x\ne 0$ and $f\left(0\right)=k$ 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.

1s1zubv

Expert

(Preamble: Apart from addition and multiplication the only genuine two-ary function allowed in print is the general power function:

The function $\left(x,y\right)↦{\mathrm{log}}_{x}y$ 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 ${\mathrm{log}}_{a}y$. 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 $0<|x|\ll 1$ one has $0<\mathrm{cos}\left(3x\right)<1$. On the other hand

Therefore the equation
${\left(\mathrm{cos}\left(3x\right)\right)}^{y}=\mathrm{cos}\left(2ix\right)=\mathrm{cosh}\left(2x\right)$
has a real solution $y$ for such $x$; and according to $\left(1\right)$ this $y$ is given by

The right hand side of $\left(2\right)$ is the correct expression of the function $f$ your teacher had in mind.
Now it comes to computing $\underset{x\to 0}{lim}f\left(x\right)$. 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.

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