Show that $\lim_{x \to +\infty}\left(f(x)+f'(x)\right)=0 \Rightarrow \lim_{x \to +\infty} f(x)=0$

Solution 1:

If $\lim\limits_{x\to+\infty} f(x) + f'(x) = 0$, then for any $\epsilon > 0$, there exists $y > 0$ such that for all $x > y$,

$$\begin{align} & f(x) + f'(x) \le \epsilon\\ \implies & \frac{d}{dx}\big[f(x) e^x\big] = (f(x)+f'(x)) e^x \le \epsilon e^x\\ \implies & f(x)e^x - f(y)e^y = \int_y^x \frac{d}{dt}\big[f(t)e^t\big]dt \le \epsilon \int_y^x e^t dt = \epsilon( e^x - e^y)\\ \implies & f(x) \le \epsilon + ( f(y) - \epsilon ) e^{y-x}\\ \end{align} $$ This leads to $$\limsup_{x\to+\infty} f(x) \le \epsilon + |f(y)-\epsilon|\lim_{x\to+\infty} e^{y-x} = \epsilon \tag{*1a}$$

Apply exactly the same argument to $-f(x)$, we get

$$\limsup_{x\to+\infty} ( -f(x) ) \le \epsilon\quad\implies\quad \liminf_{x\to+\infty} f(x) \ge -\epsilon \tag{*1b}$$

Since $\epsilon$ is arbitrary, $(*1a)$ and $(*1b)$ together implies

$$0 = \underbrace{\sup_{\epsilon>0} (-\epsilon ) \le \liminf_{x\to+\infty} f(x)}_{(*1b)} \;\le\; \underbrace{\limsup_{x\to+\infty}f(x) \le \inf_{\epsilon > 0}\epsilon}_{(*1a)} = 0 \quad\implies\quad \lim_{x\to+\infty} f(x) = 0\\ $$

Solution 2:

Hint: If, $L = \displaystyle \lim\limits_{x \to \infty} f(x) = \lim\limits_{x \to \infty}\dfrac{e^xf(x)}{e^x}$ and since $\lim\limits_{x \to \infty} e^x = \infty$, L-Hopital suggests,

$\displaystyle L = \lim\limits_{x \to \infty} \dfrac{(e^xf(x))'}{(e^x)'} = \lim\limits_{x \to \infty} f(x) + f'(x) = 0$

we still need to establish the existence of the limits !

Another way is using the Cauchy MVT.

For a $\epsilon > 0$, choose an an $\alpha > 0$, such that $\displaystyle |f(x) + f'(x)| \le \epsilon$, for $x \ge \alpha$.

Now Cauchy's Mean Value Theorem implies $\exists \beta \in (\alpha,x)$, such that

$$\displaystyle \dfrac{e^xf(x) - e^{\alpha}f({\alpha})}{e^x - e^{\alpha}} = f(\beta)+f'(\beta)$$

It follows that $|f(x) - f(a)e^{\alpha-x}| \le \epsilon|1-e^{\alpha-x}| \implies |f(x)| \le |f(\alpha)|e^{\alpha-x}+\epsilon|1-e^{\alpha-x}|$

Hence, $|f(x)| \le 2\epsilon$, for large enough $x$.

Solution 3:

This was already posted on the site, but here we go: consider the function $g$ defined by $$g(x)=\mathrm e^xf(x),$$ then $$g'(x)=\mathrm e^x(f'(x)+f(x)),$$ hence, by hypothesis, for every positive $\varepsilon$, for every $x$ large enough, say, every $x\geqslant x_\varepsilon$, $$|g(x)|\leqslant\varepsilon\mathrm e^x.$$ Integrating this on the interval $(x_\varepsilon,x)$, one gets $$|g(x)|\leqslant|g(x_\varepsilon)|+\varepsilon(\mathrm e^x-\mathrm e^{x_\varepsilon})\leqslant|g(x_\varepsilon)|+\varepsilon\mathrm e^x,$$ for every $x\geqslant x_\varepsilon$, that is, $$|f(x)|\leqslant\varepsilon+|g(x_\varepsilon)|\mathrm e^{-x}.$$ When $x\to+\infty$, this yields $$\limsup_{x\to+\infty}|f(x)|\leqslant\varepsilon.$$ This upper bound holds for every positive $\varepsilon$ hence $$\lim_{x\to+\infty}|f(x)|=0.$$