If $\lim_{x\to\infty}(f(x)+f'(x))=L$ show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$

Question

Let $f:(0,\infty) \to R$ be differentiable. Suppose that $\lim_{x\to\infty}(f(x)+f'(x))=L$. Show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$. (Hint: Write $f(x) = e^xf(x)/e^x$ and use l’Hopital’s Rule.)

My working for $\lim_{x\to\infty}f'(x)=0$:

For $\lim_{x\to\infty}f'(x) = 0$, I let $f(x) = e^xf(x)/e^x$ and applied quotient rule which then cancels off $e^{2x}$ and I'm left with $\lim_{x\to\infty}(f(x)+f'(x)-f(x))$. Can I then equate this with $\lim_{x\to\infty} \left(f(x)+f'(x)\right) - \lim_{x\to\infty}f(x)$ which then gives $L-L=0$? Is this step correct?

Answer 1

You have function $f$, and it's given that $\lim_{x \to \infty} (f(x) + f'(x)) = L.$

We'll assume that $\lim_{x \to \infty} f'(x) = 0.$

If $\lim_{x \to \infty} f'(x) = 0.$, then we can rewrite:

$$\lim_{x \to \infty} (f(x) + f'(x)) = L$$ as $$\lim_{x \to \infty} (f(x) + 0) = \lim_{x \to \infty} f(x) = L.$$

If we'll prove that $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (f(x) + f'(x))$, then we've proved that $f'(x) = 0$.

Using the hint you gave, we'll turn $f(x)$ into $\frac{e^x f(x)}{e^x}$ and using LHopital's rule that says:

$$\lim_{x \to \infty} \frac{g(x)}{h(x)} = \lim_{x \to \infty} \frac{g'(x)}{h'(x)}$$.

In our case, $$g(x) = {e^xf(x)} \text{ and } h(x) = e^x. $$

Let's start:

$$1. \text{Turn to suggested form:} \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{e^x f(x)}{e^x}.$$

$$2. \text{Apply LHopital's rule: differentiate both numerator and denumerator}:$$

$$ \lim \frac{(e^xf(x))'}{(e^x)'} = \lim = \frac{e^x f(x) + e^x f'x}{e^x} = \lim (f(x) + f'(x)).$$

We've got that $f(x) = f(x) + f'(x)$ that meets out assumation.

Therefore, $f'(x) = 0$.

Proven.

Answer 2

In case some doubt the condition of application of L'Hopital rule, we can use the discreted substitution version - i.e. Stolz theorem,which states that for two sequences $\{a_n\}$ and $\{b_n\}$, if $b_n\rightarrow\infty$ monotonously,then as long as $\lim_{n\rightarrow\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}$ exists (could be $\pm \infty$),we have \begin{align*} \lim_{n\rightarrow\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}=\lim_{n\rightarrow\infty} \frac{a_n}{b_n}. \end{align*} Now suppose $\lim_{x\rightarrow \infty} f(x)$ is not $L$,then there exists $M>0$, such that for each $X$, there exists $x>X$ such that $|f(x)-L|>M$.Choose monotonous sequence $\{x_n\}\rightarrow \infty$ such that $|f(x_n)-L|>M$.Consider $a_n=e^{x_n}(f(x_n)-L)$ and $b_n=e^{x_n}$. Then \begin{align*} \lim_{n\rightarrow\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=\lim_{n\rightarrow\infty} \frac{e^{x_n}(f(x_n)-L)-e^{x_{n-1}}(f(x_{n-1})-L)}{e^{x_n}-e^{x_{n-1}}}\\ &=\lim_{n\rightarrow\infty}\frac{e^{\xi_n}(f(\xi_n)+f'(\xi_n)-L)}{e^{\xi_n}}\\ &=\lim_{n\rightarrow\infty} f(\xi_n)+f'(\xi_n)-L\\ &=0. \end{align*} So by Stolz theorem we see \begin{align*} \lim_{n\rightarrow\infty} \frac{a_n}{b_n}&=\lim_{n\rightarrow\infty} \frac{e^{x_n}(f(x_n)-L)}{e^{x_n}}\\ &=\lim_{n\rightarrow\infty}f(x_n)-L\\ &=0, \end{align*} which contradicts that $|f(x_n)-L|$ has a non-zero lower bound. This may be equivalent to the proof of L'Hopital as the Stolz does, but I think it may resolve the doubt about where L'Hopital rule can apply.