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+1 vote

Let $x_{0}=1$ and

$x_{n+1}= \frac{3+2x_{n}}{3+x_{n}}, n\geq 0$.

$x_{\infty}=\lim_{n\rightarrow \infty}x_{n}$ is

  1. $\left(\sqrt{5}-1\right) / 2$
  2. $\left(\sqrt{5}+1\right) / 2$
  3. $\left(\sqrt{13}-1\right) / 2$
  4. $\left(-\sqrt{13}-1\right) / 2$
  5. None of the above.
asked in Calculus by Veteran (47.8k points) | 218 views

2 Answers

0 votes

Answer : C,D


as n tends to infinity,



the roots are:


Hence C,D

answered by Junior (563 points)
Not option D, because that is a negative number, and all $x_n$ are positive.
ok, got it.
I am not able to get the step after $x=1+\frac{1}{1+\frac{3}{x}}$ how we got the equation $x^{2}+x-3=0$?

x &= 1 + \frac1{1 + 3/x} = 1 + \frac1{(x + 3)/x}\\[1em]
&= 1 + \frac{x}{x+3}\\[1em]
&= \frac{(x+3) + x}{x+3}\\[1em]
x &= \frac{2x+3}{x+3}\\[1em]
x\cdot(x+3) &= 2x+3\\[1em]
x^2 + 3x &= 2x + 3\\[1em]
x^2 + x - 3 &= 0

Thank you sir. It was careless of me to overlook this simple thing.
–1 vote
clearly x(i) >= 1; now dividing numerator and denominator of x(n+1) by x(n)

x(n+1)= (3(/x(n))+2 )/(3/x(n)+1)

but as n increases, 1/(x(n) approaches 0;

therefore limit n-> infinity, x(n+1)=(0+2)/(0+1)=2.
answered by Junior (563 points)
You're saying that if $n$ is very large, $x_n$ is very large, and so $\frac1{x_n}$ tends to $0$,

But then you said that if $n$ is very large (tends to infinity), then $x_{n+1}$ is very close to $2$. This also means that $x_n$ is very close to $2$. But that contradicts your earlier statement about $x_n$ being very large.
sorry 2 is wrong

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