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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.

Any ideas? Thanks in advance.

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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.

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  • $\begingroup$ I can help you further, but first let me know if these hints makes it obvious to you now. $\endgroup$
    – RRL
    Apr 28, 2019 at 4:07
  • $\begingroup$ I still can't see how to use uniform continuity. Could you, please, explain it further? $\endgroup$ Apr 28, 2019 at 4:13
  • $\begingroup$ I shall do so... $\endgroup$
    – RRL
    Apr 28, 2019 at 4:18
  • $\begingroup$ What about the example given in the question? $\endgroup$ Apr 28, 2019 at 4:48
  • $\begingroup$ @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. $\endgroup$
    – RRL
    Apr 28, 2019 at 5:03

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