If f(x)=1/3[f(x+1) + {5/f(x+2)}] Then evaluate lim f(x) where x tends to infinity. (F(x)>0 for all x€R.
@ganeshie8 sir r u there?
Assume that the limit exists
Let \(\lim\limits_{x\to\infty}f(x) = L\)
Would you agree that Let \(\lim\limits_{x\to\infty}f(x) = \lim\limits_{x\to\infty}f(x+1) = \lim\limits_{x\to\infty}f(x+5) = L\) ?
Since the sequence \(f(n)\) converges, the value of \(f(n)\) will be close to \(L\) for all large values of \(n\) and in the limiting case, they all are equal to \(L\)
\[f(x)=1/3[f(x+1) + {5/f(x+2)}] \] take the limit both sides and get \[L=1/3[L+ 5/L] \] you can solve \(L\)
Wait please! I m not getting how u came up with limits of f(x) to b equal to that for f(x+1) and f(x+2)...
as an example, maybe consider \[f(x) = \dfrac{3x+2}{2x+1}\] \(f(x+1) = ?\) \(f(x+2) = ?\) \(\lim\limits_{x\to\infty}f(x)=?\) \(\lim\limits_{x\to\infty}f(x+1)=?\) \(\lim\limits_{x\to\infty}f(x+2)=?\)
Did u literally made up for a function that satisfies the condition of f(x) to be true? (I mean it might have took a great deal of tym) :)
Btw. Is there a theoretical way also to prove that limf(x) =Limf(x+1)?
I did not make up that function. Any function that has a finite limit satisfies the above relation
the proof is not hard but do you get the intuition on why f(x) and f(x+a) will have the same limiting value ?
Maybe bcoz if we substitute x as infinity in the f(x) we get f(infinity) and similarly x+1 would be infinity. So f(x+1) will be also f(infinity) so we can say that the two r identical.
that may do for now in order to get more sense of this, you will need to prove it rigorously using epsilon-delta definition or sequences definition of the limit
i suggest you not get into them now...
Okay! This will work for now.(what i told) Ryt?
Yes, as x -> infinity, we can say that x + 1 is same as x + 2
Ty.
yw
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