Prove / disprove question (just before Calculus exam!) So I am going to take the final exam for Calculus I.
First, I'd love if some of you have some good questions for me to solve, or some good tips and tricks that I should notice.
Second, I have a prove/disprove question that I am trying to solve, and will appreciate your help!
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function and let $L \in \mathbb{R}$.
If $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=L$, than $\lim_{x \to \infty} \frac{f(x)}{x}=\frac{L}{2}$.
So I have a hint to define a function $g(x)=x f(x)$.
So $g'(x)=f(x)+xf'(x)$.
Also, $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{f(x)}{x}+f'(x)=L$
I wanted to use L'Hopital's rule for $\lim_{x \to \infty} \frac{f(x)}{x}$, but then I have noticed that I do not know if $\lim_{x \to \infty} \frac{f(x)}{x}$ even exists.
Any hints will be appreiciated!
Thanks a lot!
Given that $$\lim \frac{g'(x)}{x}=L$$ , now assuming $f(x)\neq0$ then as $x\to\infty$ then $g(x)=xf(x)$ goes to infinity(or negative infinity) thus we can apply L'Hopital's rule , $$\lim \frac{g(x)}{\frac{x^2}2}=L\\\implies\lim \frac{2xf(x)}{x^2}=\lim \frac{2f(x)}{x}=L$$. Now if $f=0$ then the result is obvious.
You can apply L'Hopital's rule backwards and integrate, given that the limit $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{g'(x)}{x}=L$ exists:
$$L=\lim_{x \to \infty}\frac{\int g'(x)dx}{\int xdx}=\lim_{x \to \infty}\frac{g(x)}{x^2/2}=\lim_{x \to \infty}\frac{2xf(x)}{x^2}=\lim_{x \to \infty}\frac{2f(x)}{x}$$
Then, $$\lim_{x \to \infty}\frac{f(x)}{x}=\frac{L}{2}$$