<s>[INST] find properties(domain , even , surjective , injective) and asymptotes of a given function 
Given the function $f:\Bbb R \to \Bbb R$:
$
f(x)=
\begin{cases}
 \frac{\pi^2 -x^2}{x}&\text{if}\, -\infty<x<-\pi\\
 \pi^2-x^2&\text{if}\, -\pi\leq x<0\\
 1&\text{if}\, x=0\\
 \pi^2-x^2&\text{if}\, 0< x\leq \pi\\
 \frac{1}{x}\cdot \sin x &\text{if}\, \pi< x< \infty\\     
\end{cases}
$

*

*Find the domain

*Check if the function is injective , surjective and if the function is even

*Find the asymptotes in case they exist.




*

*The function is set for all real numbers since it is even defined for $x=0$ so the domain is $D=\{x \in \Bbb R\}$

*I did not understand how to approach this because there are different cases, do I approach each one alone?

If so,
Evenness: $\frac{\pi^2 -(-x)^2}{-x}\not=\frac{\pi^2 -x^2}{x}$ it is not even , but it is even in $-\pi \leq x <0$ and $0<x 
\leq \pi$ and also not even for $\frac{1}{x}\cdot \sin x$ as $\sin x$ is an odd function
Injective: for $\pi^2-x^2$ it is not injective because it is even  and $\frac{1}{x}\cdot sinx$ is not injective because $sinx$ is not injective , I am not sure but $\frac{\pi^2 -x^2}{x}$ is injective
surjective: I think it is because for every $f(x)$ we have $f(x)=y$
Asymptotes I checked every limit in the function
$\lim_\limits{x \to -\infty} \frac{\pi^2 -x^2}{x}=\infty$
$\lim_\limits{x \to -\pi^-} \frac{\pi^2 -x^2}{x}=0$
$\lim_\limits{x \to -\pi^+} \pi^2-x^2=0$
$\lim_\limits{x \to 0^-} \pi^2-x^2=\pi^2$
$\lim_\limits{x \to 0^+} \pi^2-x^2=\pi^2$
$\lim_\limits{x \to \pi^-} \pi^2-x^2=0$
$\lim_\limits{x \to \pi^+} \frac{\sin x}{x}=0$
$\lim_\limits{x \to \infty} \frac{\sin x}{x}=0$
From the last limit we can see that there is vertical asymptote  as $y=0$ (Not sure if it is considered one if $y=0$)
For oblique asymptote:
$a=\lim_\limits{x \to \infty} \frac{f(x)}{x}=\frac{\sin x}{x^2}=0$
$b=\lim_\limits{x \to \infty} f(x)-ax0$
No oblique asymptotes as well
Hopefully the translation are understandable.
I would like to know if my approach is correct, if yes then is there a different way?
If my approach is wrong then how do I approach these type of questions?
Thank you!
 [/INST] You are correct that the function is not even.  However, to show that the function is not even, you should demonstrate that there exist some $x \in \mathbb{R}$ such that $f(x) \neq f(-x)$.  For instance, $$f(2\pi) = \frac{\sin(2\pi)}{2\pi} = 0 \neq \frac{3\pi}{2} = \frac{-3\pi^2}{-2\pi} = \frac{\pi^2 - 4\pi^2}{-2\pi} = \frac{\pi^2 - (-2\pi)^2}{-2\pi} = f(-2\pi)$$
Your argument that
$$\frac{\pi^2 - x^2}{x} \neq \frac{\pi^2 - (-x)^2}{-x}$$
is not valid since the definition $f(x) = \frac{\pi^2 - x^2}{x}$ applies in the interval $(-\infty, -\pi)$ but not in the interval $(\pi, \infty)$.  Similarly, the definition $f(x) = \frac{\sin x}{x}$ applies in the interval $(\pi, \infty)$ but not in the interval $(-\infty, -\pi)$.
To show the function is not injective, you should produce two distinct elements in the domain with the same image.  For instance,
$$f\left(-\frac{\pi}{2}\right) = \pi^2 - \left(-\frac{\pi}{2}\right)^2 = \pi^2 - \frac{\pi^2}{4} = \frac{3\pi^2}{4} = \pi^2 - \frac{\pi^2}{4} = \pi^2 - \left(\frac{\pi}{2}\right)^2 = f\left(\frac{\pi}{2}\right)$$
or
$$f(2\pi) = \frac{\sin(2\pi)}{2\pi} = 0 = \frac{\sin(3\pi)}{3\pi} = f(3\pi)$$
To prove that the function is surjective, you must show that given $y \in \mathbb{R}$, there exists $x \in \mathbb{R}$ such that $f(x) = y$.  Notice that the function does not approach $-\infty$.  That suggests that it is bounded below.  In the interval $(-\infty, \pi)$,
$$f(x) = \frac{\pi^2 - x^2}{x} > 0$$
since both the numerator and denominator are negative.
In the interval $[-\pi, 0)$, $f(x) = \pi^2 - x^2 = (\pi + x)(\pi - x) \geq 0$, with equality holding if and only if $x = -\pi$ since both factors are positive if $-\pi < x < 0$ and $\pi + x = 0$ if $x = \pi$.
Clearly, $f(0) = 1 > 0$.
In the interval $(0, \pi]$, $f(x) = \pi^2 - x^2 = (\pi + x)(\pi - x) \geq 0$ since both factors are positive if $0 < x < \pi$ and $\pi - x = 0$ if $x = \pi$.
In the interval $(\pi, \infty)$,
$$f(x) = \frac{\sin x}{x} \geq \frac{\sin x}{\pi} \geq -\frac{1}{\pi}$$
since $x \geq \pi$ and $\sin x \geq -1$.
Thus, the function is bounded below by $-1/\pi$.  Hence, there is no value of $x$ in the domain such that $f(x) = -1$, so the function is not surjective.
A function has a vertical asymptote $x = a$ if $a \in \mathbb{R}$ and at least one of the following four statements is true:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty\\
\lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty\\
\lim_{x \to a^-} f(x) & = -\infty
\end{align*}
You have correctly shown that the function does not approach infinity or negative infinity at any finite value of $x$, so the function does not have a vertical asymptote.
A function has a horizontal asymptote $y = b$ if at least one of the following statements is true:
\begin{align*}
\lim_{x \to \infty} f(x) & = b\\
\lim_{x \to -\infty} f(x) & = b
\end{align*}
You have correctly shown that
$$\lim_{x \to \infty} f(x) = 0$$
Thus, the line $y = 0$ is a horizontal asymptote of the graph of $f$.
A function has an oblique asymptote $y = mx + b$, with $m \neq 0$, if at least one of the following statements is true:
\begin{align*}
\lim_{x \to \infty} [f(x) - (mx + b)] & = 0\\
\lim_{x \to -\infty} [f(x) - (mx + b)] & = 0
\end{align*}
Observe that in the interval $(-\infty, -\pi)$,
$$f(x) = \frac{\pi^2 - x^2}{x} = \frac{\pi^2}{x} - x$$
As $x \to -\infty$, the term $\pi^2/x$ approaches $0$, while the term $-x$ approaches infinity.  That suggests that the line $y = -x$ may be an oblique asymptote.  Indeed, it is since
\begin{align*}
\lim_{x \to -\infty} [f(x) - x] & = \lim_{x \to -\infty} \left[\frac{\pi^2 - x^2}{x} - (-x)\right]\\
& = \lim_{x \to -\infty} \left[\frac{\pi^2 - x^2}{x} + x\right]\\
& = \lim_{x \to -\infty} \frac{\pi^2 - x^2 + x^2}{x}\\
& = \lim_{x \to -\infty} \frac{\pi^2}{x}\\
& = 0
\end{align*}
</s>