Why is this valid? Continuity of Functions Problem

[tellmesomething]

Homework Statement

If $f(x)$ is continuous and $f(\frac{9}{2})=\frac{2}{9}$ then the value of $\lim{x\to 0}f(\frac{1-cos3x}{x²})$is?

Relevant Equations

None

I was unsure how to bring the output expression. We can assume f(x) to be f(y) for simplicity. So $y=\frac{1-cos3x}{x²}$ which is alright. Next I thought to take the limit of this expression but im unsure if my reasoning is correct.

Since the functional value is continuous so the part of the graph in the neighborhood of f(y) for the corresponding value of y (I.e limiting value of f(y)) will be equal to the actual value of f(y).
This must imply that the neighborhood of y must be in the domain of the function and hence could say that the limiting value of y=y itself..
Do I make any sense?

 

[FactChecker]

Homework Statement: If $f(x)$ is continuous and $f(\frac{9}{2})=\frac{2}{9}$ then the value of $\lim{x\to 0}f(\frac{1-cos3x}{x²})$is?
Relevant Equations: None

I was unsure how to bring the output expression. We can assume f(x) to be f(y) for simplicity. So $y=\frac{1-cos3x}{x²}$ which is alright. Next I thought to take the limit of this expression but im unsure if my reasoning is correct.

Since the functional value is continuous so the part of the graph in the neighborhood of f(y) for the corresponding value of y (I.e limiting value of f(y)) will be equal to the actual value of f(y).
This must imply that the neighborhood of y must be in the domain of the function and hence could say that the limiting value of y=y itself..
Do I make any sense?

This: $f(\frac{9}{2})=\frac{2}{9}$ will only help if $\lim \limits_{x\to 0}\frac{1-cos3x}{x²}$ is $\frac{9}{2}$. Can you continue from there?

 

[tellmesomething]

This: $f(\frac{9}{2})=\frac{2}{9}$ will only help if $\lim \limits_{x\to 0}\frac{1-cos3x}{x²}$ is $\frac{9}{2}$. Can you continue from there?

I understand that. But I want to know if $\lim\limits_{x\to 0} f( \frac{1-cos3x}{x²})=f(\lim\limits_{x\to 0}\frac{1-cos3x}{x²})$ and if so why? And where would it not be equal.

 

[renormalize]

I understand that. But I want to know if $\lim\limits_{x\to 0} f( \frac{1-cos3x}{x²})=f(\lim\limits_{x\to 0}\frac{1-cos3x}{x²})$ and if so why? And where would it not be equal.

See, for example: https://www.people.vcu.edu/~rhammack/Math200/Text/Chapter11.pdf