Jovany Clayton

Answered

2022-07-12

Find the value of K. Use of l Hospital's rule and expansion is not allowed.

Let $f(x)={\mathrm{log}}_{cos3x}(\mathrm{cos}2ix)$ 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.

Let $f(x)={\mathrm{log}}_{cos3x}(\mathrm{cos}2ix)$ 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.

Answer & Explanation

1s1zubv

Expert

2022-07-13Added 17 answers

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

$\begin{array}{}\text{(1)}& (x,y)\mapsto {x}^{y}:=\mathrm{exp}(y\phantom{\rule{thinmathspace}{0ex}}\mathrm{log}x)\phantom{\rule{2em}{0ex}}(x>0,y\in \mathbb{C})\text{}.\end{array}$

The function $(x,y)\mapsto {\mathrm{log}}_{x}y$ 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 $0<\mathrm{cos}(3x)<1$. On the other hand

$\mathrm{cos}(2ix)=\frac{{e}^{-2x}+{e}^{2x}}{2}=\mathrm{cosh}(2x)\phantom{\rule{2em}{0ex}}(x\in \mathbb{R})\text{}.$

Therefore the equation

${(\mathrm{cos}(3x))}^{y}=\mathrm{cos}(2ix)=\mathrm{cosh}(2x)$

has a real solution $y$ for such $x$; and according to $(1)$ this $y$ is given by

$\begin{array}{}\text{(2)}& y=\frac{\mathrm{log}{\textstyle (}\mathrm{cosh}(2x){\textstyle )}}{\mathrm{log}{\textstyle (}\mathrm{cos}(3x){\textstyle )}}=:f(x)\phantom{\rule{2em}{0ex}}(0<|x|\ll 1)\text{}.\end{array}$

The right hand side of $(2)$ is the correct expression of the function $f$ your teacher had in mind.

Now it comes to computing $\underset{x\to 0}{lim}f(x)$. 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.

$\begin{array}{}\text{(1)}& (x,y)\mapsto {x}^{y}:=\mathrm{exp}(y\phantom{\rule{thinmathspace}{0ex}}\mathrm{log}x)\phantom{\rule{2em}{0ex}}(x>0,y\in \mathbb{C})\text{}.\end{array}$

The function $(x,y)\mapsto {\mathrm{log}}_{x}y$ 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 $0<\mathrm{cos}(3x)<1$. On the other hand

$\mathrm{cos}(2ix)=\frac{{e}^{-2x}+{e}^{2x}}{2}=\mathrm{cosh}(2x)\phantom{\rule{2em}{0ex}}(x\in \mathbb{R})\text{}.$

Therefore the equation

${(\mathrm{cos}(3x))}^{y}=\mathrm{cos}(2ix)=\mathrm{cosh}(2x)$

has a real solution $y$ for such $x$; and according to $(1)$ this $y$ is given by

$\begin{array}{}\text{(2)}& y=\frac{\mathrm{log}{\textstyle (}\mathrm{cosh}(2x){\textstyle )}}{\mathrm{log}{\textstyle (}\mathrm{cos}(3x){\textstyle )}}=:f(x)\phantom{\rule{2em}{0ex}}(0<|x|\ll 1)\text{}.\end{array}$

The right hand side of $(2)$ is the correct expression of the function $f$ your teacher had in mind.

Now it comes to computing $\underset{x\to 0}{lim}f(x)$. 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.

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