5

Let $f(x)$ be a differentiable function and suppose $\lim_{x\to\infty}{\left(f(x+3)-f(x)\right)} = 2013$. Calculate $\lim_{x\to\infty}\frac{f(x)}{x}.$

My attempt:
The equation $\lim_{x\to\infty} f(x+3)-f(x)=2013$ suggests that the function grows linearly while $x\to \infty$, $f(x+3) - f(x) \approx f'(x) \cdot 3 = 2013 \implies f'(x) = \frac{2013}{3} = 671$. So, $f(x) = 671x + C$.
$\lim_{x \to \infty} \frac{f(x)}{x} = 671$
I'm not sure that it's correct. Can you help me with modifying my solution?

RobPratt
  • 50,938
Svyat_Smirnov
  • 61
  • 3
  • Welcome to MSE! Please note that it's not the best idea to put the whole problem in the title. – Reza Rajaei Nov 29 '24 at 22:39
  • To me it is obvious to spot that $f(x) = 671x + C$ is a family of solutions. But for the answer to be $671,$ we also need to show that if $\frac{f(x)}{x}\not\to 671,$ then $\lim_{x\to\infty}{\left(f(x+3)-f(x)\right)} \not\to 2013.$ Although this should not be too difficult. – Adam Rubinson Nov 29 '24 at 23:03
  • If you trust the problem setter, the answer to the limit must be the same for all functions that meet the requirement. @AdamRubinson has given you a family of functions that do with the common limit $671$, so if there is a consistent answer that is it. – Ross Millikan Nov 30 '24 at 01:17

0 Answers0