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Let $f$ be a continuously differentiable function on $\mathbb{R}$ such that both the limits $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x > \rightarrow \infty} f'(x)$ exist and are finite. Is $\lim_{x > \rightarrow \infty} f'(x) = 0$ necessarily?

This is true. Let $\lim_{x \rightarrow \infty} f'(x) = M$ and $\lim_{x \rightarrow \infty} f(x) = L$. Suppose $M > 0$, there exists some $A_1 \in \mathbb{R}$ such that for all $x > A_1$, $$ f'(x) > \frac{M}{2}$$ Given $\epsilon = \frac{M}{4} > 0$, there exists $A_2 \in \mathbb{R}$ such that whenever $x > A_2$, $$|f(x) - L| < \frac{M}{4}$$ For $x, y > A_2$, we have $$ |f(x) - f(y)| \leq |f(x) - L| + |f(y) - L| < \frac{M}{2}$$ Let $A = \max\,\{A_1, A_2\}$. Take $x > A$, then applying LMVT on $[x, x+1]$, there exists some $c \in (x, x+1)$ such that $$ |f(x+1) - f(x)| = |f'(c)| > \frac{M}{2}$$ But as $x > A_2$, $$|f(x + 1) - f(x)| < \frac{M}{2}$$ This is a contradiction, so $M > 0$ must be false.

Similarly, we can prove that $M < 0$ does not hold. Thus, $M = 0$.

Does this proof work? Is there a cleaner way to do this?

Anu
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$f'(x) \to l$ in $+\infty$.

If $l > 0$ then $\exists \ A$ such that $\forall x > A \ f'(x) > l/2 > 0$. So $f(x) > l/2*x + C \to +\infty$. Contradiction. The same if $l<0$.

  • Thanks! Did that last step involve using the fundamental theorem of calculus, and the property that if $f \leq g$, $\int_a^b f \leq \int_a^b g$? – Anu Feb 01 '18 at 17:21
  • Yes the fundamental theorem is used with that property. It's pretty fast like that. –  Feb 01 '18 at 17:23