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If $f'(x)>c, \forall x\in(a,+\infty)$ where $c>0$. Prove that $\lim_{x\to+\infty} f(x) = +\infty$. I would say that this is trivial, how could we prove this explicitly?

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    $\begingroup$ Hard to see why you think it's trivial, given that you can't do it. Anyway, the basic tool for proving more or less anything about derivatives is the Mean Value Theorem. $\endgroup$ – David C. Ullrich May 19 at 14:27
  • $\begingroup$ I don't know if trivial is right expression, but in my language I would say it is so obvious I could easily use it in other problems. I would say since its derivative is greater than zero, function is growing(or ascending idk which is correct) thus its limit is plus infiity, but this seems too simple to be enough... $\endgroup$ – Nigruteen May 19 at 14:35
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    $\begingroup$ You shouldn't use it in other problems if you don't know how to prove it! No, noting that $f$ is increasing is certainly not enough, for example if $f(x)=\frac x{x+1}$ then $f$ is increasing but $f$ does not tend to infinity at infinity. $\endgroup$ – David C. Ullrich May 19 at 14:41
  • $\begingroup$ @Nigruteen Perhaps mentioning your background and how much calculus you already know will help users answer your question better. $\endgroup$ – R_D May 19 at 14:44
  • $\begingroup$ @Nigruteen: Hardy mentioned about the use of "obvious" in his A Course of Pure Mathematics. As per him if a result is "obvious" then it means the reader can supply the proof very easily in his/her mind and this allows a textbook author to skip the proof in text and just mention that it is "obvious". $\endgroup$ – Paramanand Singh May 19 at 14:59
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Intuition suggests that a function with a positive derivative is strictly increasing. You can prove this using the mean value theorem. Next, you can use the mean value theorem to prove the function is also unbounded from above.

You now have a strictly increasing function with no upper bound, so you know its limit.

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This is a direct application of Newton-Leibniz formula $$ f(x) = f(a) + \int_a^x f'(t) \, dt \ge f(a) + c(x-a) $$

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  • $\begingroup$ I am first college year and we still haven't studied integrals. I would say since its derivative is greater than zero, function is growing(or ascending idk which is correct) thus its limit is plus infiity but is this correct? I'd say this is way too simple to be correct proof $\endgroup$ – Nigruteen May 19 at 14:39
  • $\begingroup$ @Nigruteen: Just having derivative positive does not mean function tends to infinity. An example is $\arctan x$. Here the hypothesis is stronger in the sense that not only derivative is positive it is also bounded away from zero (maintain at least some mimimum distance $c$ from zero). And that changes the behavior of function. $\endgroup$ – Paramanand Singh May 19 at 15:02
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    $\begingroup$ This also assumes that derivative is integrable (which may not always be the case). A better and easier approach is via mean value theorem. Mean value theorem is one of the most significant theorems of differential calculus but intro calculus texts reduce it to finding a $c$ which fits the formula $$f'(c) =\frac{f(b) - f(a)} {b-a} $$ $\endgroup$ – Paramanand Singh May 19 at 15:06
  • $\begingroup$ @ParamanandSingh I was thinking about it. I used MVT mostly for proving some inequalities. How could we use it for proving (un)boundness? I tried to figure it out I have no idea $\endgroup$ – Nigruteen May 19 at 15:16
  • $\begingroup$ @Nigruteen: we have by MVT $$f(x) =f(a) +(x-a) f'(\xi)$$ and this is greater than $f(a) +c(x-a) $ and thus tends to $\infty $ $\endgroup$ – Paramanand Singh May 19 at 15:21

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