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Imagine you know a function obeys the following: $$ \lim_{x \to \infty} f(x) = 0 $$ Assuming that is a smooth and continuous function, is is possible to prove that it also should obey this: $$ \lim_{x \to \infty} f'(x) = 0 $$ My approach: $$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ \lim_{x \to \infty}f'(x) = \lim_{x \to \infty} \left (\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \right) \\ \lim_{x \to \infty}f'(x) = \lim_{h \to 0} \frac{\lim_{x \to \infty} f(x + h) - \lim_{x \to \infty} f(x)}{h} \\ \lim_{x \to \infty}f'(x) = \lim_{h \to 0} \frac{0 - 0}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \\ $$ Is this reasoning correct? If not, what is wrong? Are there other ways to prove this?

Álvaro Rodrigo
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    I do not think it is true so it cannot be proved. Try finding the derivative of $f(x)=\dfrac{\sin(x^3)}{x}$ for $x>1$ and consider what values the function and its derivative can achieve – Henry Jan 02 '23 at 00:13
  • Or even $x \mapsto {\sin x^2 \over x}$. – copper.hat Jan 02 '23 at 00:18
  • @copper.hat - indeed, but my version has unbounded derivatives as $x$ increases – Henry Jan 02 '23 at 00:19
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    You can found a counterexample in other MSE questions, e.g., MSE 162078 – Benjamin Dickman Jan 02 '23 at 00:20
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    For completeness, the error in your proof lies in the fact that, in general, you cannot interchange the order of double limits, i.e you cannot say that $\lim_{x\to x_0}\lim_{y\to y_0} g(x, y) = \lim_{y\to y_0}\lim_{x\to x_0} g(x,y)$. – Giorgos Giapitzakis Jan 02 '23 at 00:28
  • @MyMathYourMath Don’t you mean to interchange function and derivative? But, even so, why is this relevant? – Ted Shifrin Jan 02 '23 at 02:17

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