0
$\begingroup$

Suppose we have a function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined recursively by $$f(x)=\frac{1}{3}\left[f(x+1)+\frac{5}{f(x+2)}\right] $$

Assuming $~f(x) >0$ for all $x\in\mathbb{R}$, compute $\lim_{x\rightarrow\infty}f(x)$

  • 0
    Yes , and the answer is √5/2 – 2017-02-14

2 Answers 2

2

Assuming $\lim_{x\rightarrow\infty}f(x)$ exists and is finite, we have $$ \lim_{x\rightarrow\infty} f(x)=\frac{1}{3}\left[\lim_{x\rightarrow\infty} f(x+1)+\frac{5}{\lim_{x\rightarrow\infty}f(x+2) }\right]. $$ Since $\lim_{x\rightarrow\infty}f(x+c) =\lim_{x\rightarrow\infty}f(x)$ for all $c\in\mathbb{R}$ (since $c$ is negligible with respect to $\infty$), we have $$ \lim_{x\rightarrow\infty} f(x)=\frac{1}{3}\left[\lim_{x\rightarrow\infty} f(x)+\frac{5}{\lim_{x\rightarrow\infty}f(x) }\right]. $$

Let $a:=\lim_{x\rightarrow\infty} f(x)$, the above equation yields $$ 3a=a+\frac{5}{a}, $$ which gives $a=\pm\sqrt{\frac{5}{2}}$ as a result. We know that $f(x)>0$ for all $x$, so we conclude $a=+\sqrt{\frac{5}{2}}$ which is what we expected.

  • 0
    f(x) >0 has nothing to do with the question then? – 2017-02-14
  • 0
    Actually it does play a role, sorry, I slightly edited my answer to include that part – 2017-02-14
  • 0
    Can u tell how to use mathematical notations? – 2017-02-14
  • 0
    you mean when formatting the text? You can use standard LaTeX notation. Look at the code of my answer to get an example. – 2017-02-14
  • 0
    By the way, are you satisfied with the answer? – 2017-02-14
  • 0
    I think I am but how to get that intuition to slove it like that ? – 2017-02-14
  • 0
    Well, the key point is just that when you let $x$ tend to infinity, every constant you add to $x$ is simply irrelevant, since it is extremely small compared to infinity. So in this case you have $\lim_{x\rightarrow\infty}f(x+2)=\lim_{x\rightarrow\infty}f(x+1)=\lim_{x\rightarrow\infty}f(x)$ – 2017-02-14
  • 0
    I don't have enough reputation to upvote or accept it I guess – 2017-02-15
  • 0
    What? You asked the question! You can definitely accept... No reputation required to accept answers to your own questions! You are actually the only one who can accept an answer... – 2017-02-15
  • 0
    When I tap the upward arrow beside Ur answer. This is popping (Thanks for the feedback! Votes cast by those with less than 15 reputation are recorded, but do not change the publicly displayed post score.) – 2017-02-15
  • 0
    Ok you cannot upvote, but you can certainly click the check mark symbol ✓ close to the answer to accept it – 2017-02-15
  • 0
    The one problem with this is that it doesn't address the question of whether $\lim_{x\to\infty} f(x)$ exists in the first place. If the limit exists, then the calculation is correct and tells you what it must be. But if the limit does not exist, then the calculation is meaningless and tells you absolutely nothing. – 2017-02-15
  • 0
    @PaulSinclair You are right. Although I doubt that OP is interested in the case where the limit does not exist (since he clearly stated in the comments that the answer was $\sqrt{5/2}$), I have slightly changed my answer to include that possibility. – 2017-02-15
  • 0
    Yes, but I always prefer to point out the limitation of this technique to the uninitiated, because it is an easy pitfall that people regularly fall into. There are infinitely (uncountably) many functions $f$ that satisfy that formula. But I suspect that only for the constant function $f(x) = \sqrt{5/2}$ does the limit converge.. – 2017-02-15
  • 0
    Agreed. Thanks for pointing it out – 2017-02-15
  • 0
    @Rohit sorry if I ask again. Since you are satisfied with the answer, would you mind clicking the ✓ symbol to accept it? – 2017-02-17
  • 0
    Great answer man , really nicely explained. Had my doubt satisfactorily resolved. :) – 2017-12-13
  • 0
    @Tanuj Thanks! I really appreciate it :) – 2018-01-20
0

Let $L$ denote this limit, assuming it exists. Then taking the limit of both sides of your equation, $$L = \frac{1}{3}(L + \frac{5}{L}).$$ Using this data and the fact that $f(x) > 0$ for every $x$, we can use algebra to solve for $L$.