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If $x_1=1$ and $x_{n+1}=\large\frac{4+3x_n}{3+2x_n}\qquad$$n \geq 1$ and if $\lim\limits_{n\to \infty}x_n=l$ then l is

$(a)\;-\sqrt{2}\qquad(b)\;\sqrt 2\qquad(c)\;2\qquad(d)\;None\;of\;these$

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$x_{n+1}=\large\frac{4+3x_n}{3+2x_n}$
$\Rightarrow \lim\limits_{n\to \infty}x_{n+1}=\large\frac{4+3\lim\limits_{x\to \infty}x_n}{3+2\lim\limits_{n\to \infty}x_n}$
$\Rightarrow l=\large\frac{4+3l}{3+2l}$
$\Rightarrow l^2=2$
$\Rightarrow l=\sqrt 2$
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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