Well, how about this:
Take and such that if then and if then (these 's exist by your limit conditions) then by the Intermediate value theorem we have that for every there exists a such that . Since this is true for any we get the result.
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 is a continuous real-valued function with domain . Assume and . Prove that the range of is .
Proof: Assume the hypotheses. Since f is continuous at every point in its domain, we have that for all . Then by the definition of limit we have that for all > 0, there exists a > 0 such that whenever . That is, . So if we let , then and so . Also, if we let , then and so . Then , so the range of f is .
Again, I know it seems sketchy, so please help me identify what I need to say to make this more precise.