Find the limit of $x_{n+1} = 4 - \frac{1}{x_n}$, given that it is convergent.

Question

I already have that it is bounded above and below by 3,4 and have proven it. I just do not know how to verify that 4 is definitely the limit. Is it being bounded enough to justify the claim and if not how do I go about proving it?

Answer 1

If it converges, then $X=\lim_{n\to\infty}x_n$, and

$$X=4-\frac1X\implies X^2-4X+1=0\implies X=2\pm\sqrt3$$

Since $3\le X\le4$, we have

$$X=2+\sqrt3$$

Answer 2

Hint:

Call $x$ the limit.

Then, $$\lim_{n\to\infty} x_{n+1} = x$$

Now, what is $$\lim_{n\to\infty} \left[4-\frac1{x_n}\right]$$ equal to?

Answer 3

If the limit exists, one index shift will make little difference; so treat $$x_{n+1}=x_{n}=L$$ And solve $$ L=4-1/L $$