Limit of a function

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. If:

$$\lim_{x \rightarrow +\infty} [f(x)+f'(x)] =0$$

holds, then prove that $\displaystyle \lim_{x \rightarrow +\infty} f(x)=0$.

Solution


We are splitting the proof in several steps:

(a) Let $\epsilon>0$. Since $\displaystyle \lim_{x \rightarrow +\infty} [f(x)+f'(x)] =0$ this means that there exists a $A$ such that forall $x>A$ to hold:

$$\left | f(x)+f'(x) \right |< \frac{\epsilon}{2}$$

(b) Let $x>A$. Considering $g(x)=e^x f(x)$ then from the Cauchy Mean Value Theorem for the functions $y=g(x), \; y=e^x$ there exists a $c \in (A, x)$ such that:

$$\frac{g(x)-g(A)}{e^x-e^A}= \frac{g'(c)}{e^c}= f(c)+f'(c)$$

Hence:

$$ \left | g(x) -g(A)\right |=\left | f(c)+f'(c) \right |\left | e^x-e^A \right |< \frac{\epsilon}{2}\left | e^x-e^A \right |$$

(c) From (b) it follows that $\displaystyle \left | g(x) \right |< \frac{\epsilon}{2}\left | e^x-e^A \right |+ \left | g(A) \right |$. Therefore:

$$\begin{align*}
\left | f(x) \right | &<e^{-x}\left [ \frac{\epsilon}{2}\left | e^x-e^A \right |+\left | f(A)e^A \right | \right ] \\
 &=\frac{\epsilon}{2}\left | 1-e^{A-x} \right |+ \left | f(A)e^{A-x} \right | \\
 &=\frac{\epsilon}{2}\left ( 1-e^{A-x} \right )+ \left | f(A)e^{A-x} \right | \\
 &< \frac{\epsilon}{2}+ \left | f(A)e^{A-x} \right |
\end{align*}$$

(d) Since $\displaystyle \lim_{x \rightarrow +\infty} \left | f(A)e^{A-x} \right |=0$ (trivially) then there exists a $B$ such that:

$$\left | f(A)e^{A-x} \right |< \frac{\epsilon}{2}, \;\;\;\;\; \text{for all} \;\; x>B$$

Setting $\delta= \max \{A, B \}$ then forall $x>\delta$ we have that:

$$\left| f(x) \right| <\frac{\epsilon}{2}+\frac{\epsilon}{2}= \epsilon$$

ending the proof.