Limit of Derivative at Infinity

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and $$ \lim_{x\to\infty}f'(x)=L, $$ where $L$ is finite. We will prove that $$ \lim_{x\to\infty}\frac{f(x)}{x}=L. $$ **Step 1: Apply the Mean Value Theorem.** Fix any $a\in\mathbb R$. For every $x>a$, the Mean Value Theorem guarantees the existence of a point $c_x\in(a,x)$ such that $$ \frac{f(x)-f(a)}{x-a}=f'(c_x). $$ Since $c_x\in(a,x)$ and $x\to\infty$, we have $c_x\to\infty$ in the sense that, for every sufficiently large lower bound, all possible $c_x$ eventually exceed that bound. Therefore, $$ \lim_{x\to\infty}f'(c_x)=L. $$ Consequently, $$ \lim_{x\to\infty}\frac{f(x)-f(a)}{x-a}=L. $$ **Step 2: Relate this quotient to $\frac{f(x)}{x}$.** For $x\ne0$, $$ \frac{f(x)}{x} = \frac{x-a}{x}\cdot\frac{f(x)-f(a)}{x-a} +\frac{f(a)}{x}. $$ As $x\to\infty$, $$ \frac{x-a}{x}\to1 \qquad\text{and}\qquad \frac{f(a)}{x}\to0. $$ Using the limit obtained in Step 1, we conclude that $$ \lim_{x\to\infty}\frac{f(x)}{x}=1\cdot L+0=L. $$ Hence, $$ \boxed{\lim_{x\to\infty}\frac{f(x)}{x}=L}. $$ **Verification of the options** **Option A is false.** The derivative limit does not imply that $f(x)$ has a finite limit. For example, $f(x)=x$ gives $f'(x)=1$, but $f(x)\to\infty$. **Option B is false.** If $f(x)=5$, then $f'(x)=0$ and $\lim_{x\to\infty}f(x)=5\ne0$. **Option C is true.** As proved above, the Mean Value Theorem implies that $\displaystyle\lim_{x\to\infty}\frac{f(x)}{x}=L$. **Option D is false.** Although $f'(x)$ approaches $L$, it need not have a fixed sign when $L=0$. For example, $$ f(x)=\sin(\log(1+x^2)) $$ is differentiable on $\mathbb R$ and $$ f'(x)=\frac{2x}{1+x^2}\cos(\log(1+x^2))\to0 $$ as $x\to\infty$, but the function continues to oscillate and is not eventually monotone. Therefore, the correct option is $$ \boxed{\mathrm{C}}. $$