About a real, differentiable function with an oblique asymptote.

Question

Consider a real, differentiable function f with an oblique asymptote, i.e. $lim_{x\to\infty} f(x)-ax-b=0$. Does this imply that $lim_{x\to\infty} f'(x)=a?$

It makes intuitive sense, but coming up with a proof in terms of epsilon-delta definitions eludes me.

Is there a theorem on this, or a proof one of you could come up with?

This counterexample answers your question: Construction of function with function going to 0 but derivative not as X tends to infinity E.g. $f(x)=ax+b+\frac{\sin(x^2)}x.$ — Anne Bauval, May 15 '23 at 11:50
Thank you very much. Do you have any suggestions on how to find counter examples to things I conjecture like this? — Aqeel, May 15 '23 at 12:03

score 0 · Answer 1 · answered May 16 '23 at 08:40

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$\lim_{x\rightarrow \infty}\frac{f(x)}{x} =0$. Hospital rule does't ask your question?

answered May 16 '23 at 08:40

lib

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About a real, differentiable function with an oblique asymptote.

1 Answers1