Inverse L'hospital.... does it exist?

Question

Say we have a function $f:\mathbb{R}\to\mathbb{R}$, differentiable such that $\lim_{x\to\infty}f(x)=a, a\in\mathbb{R}$ and $\exists\lim_{x\to\infty}xf'(x).$ And I have to calculate: $$\lim_{x\to\infty}xf'(x)$$

...First let's rewrite this as:

$\lim_{x\to\infty}\frac{f'(x)}{\frac 1x}$. Now... Can I use "inverse l'hospital"? and what I mean by that is... to integrate the top and bottom and have: $\lim_{x\to\infty}\frac{f(x)}{ln(x)}$ and use the fact that $\lim_{x\to\infty}f(x)=a$.... so my limit would be $\lim_{x\to\infty}\frac {a + c_1}{ln(x)+c_2}=0?$

And if I can't do this, how can I solve this problem?

Answer 1

I am not convinced about the inverse of L'Hopital's rule at all. All I know is there are plenty of examples when it does not work.

Instead, you can use L'Hopital itself. The idea is that $(xf(x))' = f(x) + xf'(x)$.

Note that since $\lim_{x \to \infty} f(x) = a$ exists and $\lim_{x \to \infty} xf'(x)$ exists, we get that $\lim_{x \to \infty} (xf(x))'$ exists.

Now, all we need is the following transformation: $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{xf(x)}{x} = \lim_{x \to \infty} \frac{(xf(x))'}{1} = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} xf'(x) $$

Where weare a slight generalization of L'Hopital(where the indeterminate form can be anything over infinity) by the fact that the denominator is infinite at infinity, and both numerator and denominator are at least differentiable in $x > R$ for some $R$, and that $\lim_{x \to \infty} (xf(x))'$ exists, since otherwise the inequality is not valid. That is why we need $\lim_{x \to \infty} xf'(x)$ to exist before knowing its value : L'Hopital cannot be used here otherwise.

From here, it is fairly clear that $\lim_{x\to \infty} xf'(x) = 0$. I will attach a link to the other L'Hopital rule shortly.

EDIT : In this answer :If $f'$ tends to a positive limit as $x$ approaches infinity, then $f$ approaches infinity you may find the generalized rule.

Answer 2

Theorem: $\lim\limits_{x \rightarrow \rho}\frac{f_1(x)}{ f_2(x) } \stackrel{{\infty/\infty}}{=} \lim\limits_{x \rightarrow \rho}\frac{ F_1(x)}{ F_2(x) } $

Proof: Let $f_1(x)$ and $f_2(x)$ have poles at $\rho$, such $\lim\limits_{x \rightarrow \rho}\frac{f_1(x)}{f_2(x)} \stackrel{{\infty/\infty}}{=} L$. Let's notice that;

$$\forall_{f_{1,2}(x+\xi):\lim\limits_{x \rightarrow \rho} f_{1,2}(x+\xi)=L_{1,2} }\lim\limits_{x \rightarrow \rho}\frac{f_1(x)+f_1(x+\xi)}{f_2(x)+f_2(x+\xi)}=\lim\limits_{x \rightarrow \rho}\frac{f_1(x)}{f_2(x)}=L$$

In such case we can obtain;

$$ \begin{split} \lim\limits_{x \rightarrow \rho}\frac{f_1(x)}{f_2(x)}& =\lim\limits_{x \rightarrow \rho}\frac{ \frac{a-x}{|\rho-x|}f_1(x)}{\frac{a-x}{|\rho-x|}f_2(x)}\\ & =\lim\limits_{x \rightarrow \rho}\frac{\frac{a-x}{|\rho-x|}\sum_{i=0}^{n=|\frac{1}{\rho-x}|} f_1\left(x+i\frac{a-x}{|\rho-x|}\right)}{\frac{a-x}{|\rho-x|}\sum_{i=0}^{n=|\frac{1}{\rho-x}|} f_2\left(x+i\frac{a-x}{|\rho-x|}\right) } \\ & =\lim\limits_{x \rightarrow \rho}\frac{\int_x^af_1(x) dx}{\int_x^a f_2(x) dx} \end{split} $$

At this point we can set any $a$, such $|F_{1,2}(a)| \neq \infty $. Second less formal way of looking at this is to set $a \rightarrow \infty$ and assume it's convergent to $0$, which is sufficient condition to get analytic continuation of $\int_x^{\infty} f_{1,2}(x)dx$.

Let's notice that it do not work for $\frac{0}{0}$