## badreferences 3 years ago Bored?$f:\mathbb R\to\mathbb R\mid e^{x-\int f\left(x\right)}=\int f\left(x\right)$Find:$\lim_{x\to\infty}\left(f\left(x\right)\right)^x$

The answer is 0. And I'm still bored -_-

Haha! I'm right aren't I?

No, lol, I hope that isn't a real question. XD

-_- You are clearly misunderstanding me <-- (see what I did there?) The answer is F.

5. satellite73

i don't even understand the question

Also, I should've specified a $$dx$$ in the integrals, so it's clear they aren't an operation of $$f$$. But I think you all knew that already.

I didn't. So you misunderstood me again :D

8. FoolForMath

what is $$dx$$? :P

We're given conditions $$f:\mathbb R\to\mathbb R$$ which means the function maps reals into reals. We know that $$\int f\left(x\right)\,dx=e^{x-\int f\left(x\right)\,dx}$$. We want to find the limit asked knowing this much.

Could you repeat that?

@LifeIsADangerousGame You are a snarky one, aren't you?

It's my job. It's what I do.

13. Ishaan94

Solution? ಠ_ಠ

14. Ishaan94

@Mr.Math SAVVEE MEE!

15. Mr.Math

@Ishaan94 I will come back to it later and see if I can do it. I feel tired now.

16. Ishaan94

$\mathsf{ \color{yellowgreen}{\text{Okay}}}$

17. Mr.Math

By taking ln of both sides we have $x-\int f(x)dx=\ln(\int f(x)dx)) \implies 1-f(x)=\frac{f(x)}{\int f(x)dx}$ $\large \implies f(x)=\frac{\int f(x)dx}{\int f(x)dx+1}.$ Because $$\int f(x)dx>0$$, we have $$0<f(x)<1$$. Hence $\large \lim_{x\to \infty} (f(x))^x=0.$

18. Ishaan94

19. karatechopper

ISHAAN GET IN CHAT IF U R NOT HELPIN...

20. karatechopper

:D intrestin lookin R u got there

21. Mr.Math

Well, he/she should post the answer then.

22. Ishaan94

Lol, I actually mistyped the question. I think the answer for this is actually $$0$$, my bad.

24. Mr.Math

What is the question then?