# Thread: Proof that the range of f is all of R

1. ## Proof that the range of f is all of R

Hey all, I need some help with this proof for my analysis class. I have a proof that I wrote up but I'm not sure if it's correct. I'll write the claim and my idea of a proof below and if anyone could tell me what's wrong with my logic or if there's a better way to write the proof, I'd really appreciate it. I'm just not sure how to go about this so what I'm providing is my best guess.

Claim: Assume $f$ is a continuous real-valued function with domain $(-\infty , \infty )$. Assume $\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow - \infty} f(x) = - \infty$. Prove that the range of $f$ is $(-\infty , \infty )$.

Proof: Assume the hypotheses. Since f is continuous at every point in its domain, we have that $\lim_{x \rightarrow x_0} f(x) = f(x_0)$ for all $x_0 \in (-\infty , \infty )$. Then by the definition of limit we have that for all $\varepsilon$ > 0, there exists a $\delta$ > 0 such that $|f(x) - f(x_0)| < \varepsilon$ whenever $|x - x_0| < \delta$. That is, $f(x_0) - \varepsilon < f(x) < f(x_0) + \varepsilon$. So if we let $x_0 \rightarrow \infty$, then $f(x_0) \rightarrow \infty$ and so $f(x) < \infty$. Also, if we let $x_0 \rightarrow -\infty$, then $f(x_0) \rightarrow -\infty$ and so $-\infty < f(x)$. Then $-\infty < f(x) < \infty$, so the range of f is $(-\infty , \infty )$.

Again, I know it seems sketchy, so please help me identify what I need to say to make this more precise.

Take $n\in \mathbb{N}$ and $N_1,N_2$ such that if $x then $f(x)\leq -n$ and if $x>N_2$ then $f(x)\geq n$ (these $N$'s exist by your limit conditions) then by the Intermediate value theorem we have that for every $y\in (-n,n)$ there exists a $z\in \mathbb{R}$ such that $f(z)=y$. Since this is true for any $n$ we get the result.
Take $n\in \mathbb{N}$ and $N_1,N_2$ such that if $x then $f(x)\leq -n$ and if $x>N_2$ then $f(x)\geq n$ (these $N$'s exist by your limit conditions) then by the Intermediate value theorem we have that for every $y\in (-n,n)$ there exists a $z\in \mathbb{R}$ such that $f(z)=y$. Since this is true for any $n$ we get the result.