$f :\Bbb R\rightarrow \Bbb R$ is continuous differentiable with $|f′(x)| \leq (x^2 + 1)^{−1}$ for all x ∈ R. Show that $\lim $ x→+∞ $f(x)$ exists.

If $f :\Bbb R\rightarrow \Bbb R$ is continuous differentiable with $|f′(x)| \leq (x^2 + 1)^{−1}$ for all $x \in\mathbb R.$ Show that $\lim_{ x\to\infty} f(x)$ exists.

Hi guys, So I've been trying to solve this problem. I thought that I could use that $\lim_{x\to\infty} |f’(x)|=0$ to then show that $\lim_{x\to\infty} f(x)$ exists, which I thought made intuitively kinda sense until I discovered some counter examples. Now, because I’m stuck I really would appreciate if someone could give me a hint and/or point me to a concept which could help to solve the exercise. Thanks in advance!

Proving the existence of a limit without knowledge of the actual value of the limit usually involves the completeness of the underlying space, i.e. using that Cauchy sequences are convergent. Here one can proceed as follows:

• First note that for $0 < R < a < b$ $$ |f(b) - f(a)| \le \int_a^b |f'(t) \, dt \le \int_a^b \frac{dt}{1+t^2} \\ = \arctan(b)-\arctan(a) \le \frac \pi 2 - \arctan(R) $$ and the right-hand side becomes arbitrarily small for sufficiently large $R$.

• Conclude that $(f(n))_n$ is a Cauchy sequence, so that $L = \lim_{n \to \infty} f(n)$ exists.

• Finally conclude that $\lim_{x \to +\infty} f(x) = L$.