# Uniformly continuous derivative implies existence of limit

Let $$f \in C^1([0, +\infty))$$. Suppose that $$\lim_{x \rightarrow +\infty} f(x)=L$$ and $$f'$$ is uniformly continuous.

Show that $$\lim_{x \rightarrow +\infty} f'(x) + f(x)=L$$

I tried to apply L'Hospital's Rule to $$\frac{e^xf(x)}{e^x}$$ since $$\frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$$. It seems alright but I didn't use the uniform continuity of $$f'$$ and it doesn't work for the function $$f(x)=\frac{\sin(x^2)}{x}$$ whose derivative is $$f'(x)=2\cos(x^2)-\frac{\sin(x^2)}{x^2}$$ since $$\lim_{x \rightarrow +\infty} f'(x)$$ doesn't exist.

We have $$\lim_{x \to \infty} f'(x) = 0$$ because,

$$\int_0^x f'(t) \, dt = f(x) - f(0), \\\int_0^\infty f'(t) \, dt = \lim_{x \to \infty}f(x) - f(0) = L - f(0) \quad (\text{convergent})$$

and $$f'$$ is uniformly continuous.

To prove this assume that $$\lim_{x \to \infty}f'(x) =0$$ does not hold and arrive at contradiction with the fact that the integral of $$f'$$ is convergent.

If $$\lim_{x \to \infty} f'(x) = 0$$ does not hold then there exists $$\epsilon_0 > 0$$ and a sequence $$x_n \to \infty$$ such that $$|f'(x_n)| \geqslant \epsilon_0$$ for all $$n$$. Next apply uniform continuity.

Assume WLOG that $$f'(x_n) \geqslant \epsilon_0$$.

There exists by uniform continuity $$\delta > 0$$ such that $$|f'(t) - f'(x_n)| < \epsilon_0/2 \implies f'(t) > \epsilon_0/2$$ for all $$t \in [x_n - \delta,x_n + \delta],$$ and

$$\int_{x_n - \delta}^{x_n + \delta} f'(t) \, dt > \epsilon\delta$$

This violates the Cauchy criterion for convergence of the improper integral since $$x_n$$ can be arbitrarily large.

• I can help you further, but first let me know if these hints makes it obvious to you now.
– RRL
Apr 28, 2019 at 4:07
• I still can't see how to use uniform continuity. Could you, please, explain it further? Apr 28, 2019 at 4:13
• I shall do so...
– RRL
Apr 28, 2019 at 4:18
• What about the example given in the question? Apr 28, 2019 at 4:48
• @JensSchwaiger: $\cos(x^2)$ is not uniformly continuous on $[0,\infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
– RRL
Apr 28, 2019 at 5:03