Calculus Problem by LOGA 7

(A) is false, with $f(x) = \sqrt{x}$ giving a counterexample.

By the Mean Value Theorem, $f(x+1) - f(x) = f'(y_x)$ for some $x < y_x < x+1$ . Thus $\lim_{x \to \infty}[f(x+1)-f(x)] \; = \; \lim_{y \to \infty}f'(y) = 0$ . Thus (B) is true.

Consider the function $f(x) \; =\; \left\{ \begin{array}{lll} 16n^3\big(n + \tfrac{1}{2n} - x\big)^2\big(x - n + \tfrac{1}{2n}\big)^2 & \hspace{1cm} & n - \tfrac{1}{2n} < x < n + \tfrac{1}{2n}\;,\; n \in \mathbb{N} \\[2ex] 0 & & \textrm{o.w.} \end{array}\right.$

This function is differentiable everywhere. In the interval $\big(n - \tfrac{1}{2n},n+\tfrac{1}{2n}\big)$ , the function never exceeds $n^{-1}$ . Thus it is clear that $\lim_{x \to \infty}f(x) = 0$ . But the Mean Value Theorem tells us that there exists $n < v_n < n + \tfrac{1}{2n}$ such that $f'(v_n) \; = \; 2n\big[f\big(n + \tfrac{1}{2n}\big) - f(n)\big] = -2$ for all $n \in \mathbb{N}$ . Since $v_n \to \infty$ as $n \to \infty$ , we cannot have $\lim_{x \to \infty}f'(x) = 0$ . Thus (C) is false.

These are what I've found for counterexamples:

A. $f(x)=\ln x$

C. $f(x)=e^{-x}\sin(e^x)$

Quick and easy check is by utilizing $f(x) = \ln(x).$