# How to evaluate limits [duplicate]

Let $f$ be a continuously differentiable function on $\mathbb R$. Suppose that $L=\lim\limits_{x\to \infty}(f(x)+f^{'}(x))$ exists. If $0<L<\infty$, and if $\lim\limits_{x\to \infty} f^{'}(x)$ exists then, can it be proved that $\lim\limits_{x\to \infty} f^{'}(x)=0$?

Let $(x_n)$ be a sequence diverges to $\infty$. Then $f(x_n)+f^{'}(x_n)\to L$. After that I could not proceed. Please help!

• I think that the intuition behind that would be that if $f(x)$ approaches a value, then $f'(x)$ cannot approach anything but 0 – SalmonKiller May 29 '15 at 16:40

In broad strokes, the ideas are:

Since $\lim_{x\to\infty}f'(x)$ and $\lim_{x\to\infty}(f(x)+f'(x))$ exist and are finite, then $\lim_{x\to\infty}f(x)$ exists and is finite. This implies $\lim_{x\to\infty}f'(x)=0$.

• $(\sin x^2)/x$ has a limit but its derivative doesn't. – Empy2 May 29 '15 at 16:48
• @Michael, from the hypothesis we know that the derivative has a limit. What Tim is saying is that given the limit exists, it must be zero. – Joel May 29 '15 at 16:49

If $\lim_{x\to \infty} f'(x) = C \neq 0$ then there is a point $x_0$ for which $$|f'(x) - C| < C/4$$ for all $x > x_0$. Without loss of generality assume that $C>0$. Hence $$f(x) = f(x_0) + \int_{x_0}^x f'(x) dx > f(x_0) + \int_{x_0}^x \frac34 C dx = f(x_0) + \frac34 C (x-x_0).$$

Since $\lim_{x\to\infty} ( f(x) + f'(x) ) = L$ exists and $\lim_{x\to\infty} f'(x)$ also exists, this means that $\lim_{x\to\infty} f(x)$ exists. Also since the limits are both finite, so is the limit of $f(x)$.

However, from our above inequality we have $\lim_{x\to\infty} f(x) = \infty$. This is a contradiction.

• Another approach would be to notice that $f(x+1) - f(x) \to 0$ as $x\to \infty$ and then apply the mean value theorem. – Joel May 29 '15 at 16:47
• – Idris Addou May 29 '15 at 18:55