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Asymptotic Differentiable Function

Question

Let $f$ be any function asymptotic at zero. Prove/Disprove:
If $f$ is differentiable everywhere in its domain, then $\lim_{x\to \infty}f\;'(x)$ must be equal to zero?

Answer

Consider the function \[ f(x)=\frac{\sin(x^3)+2}{x^2}. \] Clearly $f$ differentiable everywhere in its domain and has an asymptote at zero. However, when we differentiate $f$ we get \[ f\;'(x)=3\cos(x^3)- 2\cdot \frac{2+\sin(x^3)}{x^3}. \] Obviously \[ \displaystyle\lim_{x\to\infty} 2\cdot \frac{2+\sin(x^3)}{x^3}=0. \] Thus we can use Arithmetics of limits to conclude that: If $\lim_{x\to\infty} f\;'(x)=0$, then $\lim_{x\to 0} 3\cos(x^3)=0$. However, $\lim_{x\to\infty}3\cos(x^3)$ does not exist, to see this consider the limit of the sequence \[ 3\cdot \cos\left(\left(\sqrt[3]{n\pi}\right)^3\right). \] Since \[ \displaystyle\lim_{n\to\infty} 3\cdot \cos\left(\left(\sqrt[3]{2n\pi}\right)^3\right)=3 \qquad \mbox{and} \qquad \displaystyle\lim_{n\to\infty} 3\cdot \cos\left(\left(\sqrt[3]{(2n+1)\pi}\right)^3\right)=-3 \] the limit \[\lim_{n\to\infty} 3\cdot \cos\left(\left(\sqrt[3]{n\pi}\right)^3\right)\] does not exist. This prove that, $\lim_{x\to\infty} f\;'(x)$ does not exist as well. ■