Prove that the function is not injective - Calculus problem I am stuck with this problem and I'll appreciate any hint on how to continue - 
Let $f:\mathbb R\to\mathbb R$ be a continuous function, with  $\lim_{x\to\infty}⁡f(x)= \lim_{x\to-\infty}⁡f(x)=0$.
Prove that f is not an injective function.
I thought of proving be contradiction and using the the fact that an injective and continuous function is strictly monotonic. I believe it is a right approach be I don't know how to continue.
Thank you! 
 A: It can't be injective.
If $f(x)=0$ for all $x$, then clearly it is not injective.
Suppose there exists $a$ such that $f(a)\neq 0$. Let $\epsilon=\frac{|f(a)|}{2}$. Since $\lim_{x\to\infty}f(x)=0$, there exists $M\gt 0$ such that if $x\gt M$, then $|f(x)|\lt\epsilon$. And since $\lim_{x\to-\infty}f(x) = 0$, there exists $N\gt 0$ such that if $x\lt -N$, then $|f(x)|\lt\epsilon$. Note that $-N\leq a\leq M$ must hold.
Now let $b=f(M+1)$ and $c=f(-N-1)$. Note that $|b|\lt |f(a)|$, $|c|\lt |f(a)|$. Thus, you can find a value $d$ that is both strictly between $b$ and $f(a)$, and also strictly between $c$ and $f(a)$. Using the intermediate Value Theorem, you can find a point $x_1$ between $a$ and $M+1$ and a point $x_2$ between $-N-1$ and $a$ but with $f(x_1)=f(x_2)$.
A: One can also argue as suggested in the question. Assume $f$ is injective; then $f$ is either increasing or decreasing (that's where we use continuity). Assume increasing. If $\lim_{x\to-\infty}f(x)=0$, as $f$ is increasing, we have $f(0)>0$ (any other point in the line will do). And $f(x)>f(0)$ for all $x>0$, making it impossible that $\lim_{x\to\infty}f(x)=0$. If $f$ is decreasing, we'll get $f(0)<0$ and we apply a similar reasoning. 
If instead we assume $\lim_{x\to\infty}f(x)=0$, we work with $f(-x)$ and apply the previous paragraph. In the end, we have proven that if $f$ is injective it is impossible to have both $\lim_{x\to\pm\infty}f(x)=0$. Thus, if both limits are zero then $f$ is not injective. 
