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enter image description hereenter image description here In the solution , I think there is a mistake the value of l should be π and 1-π it should not be π+1. Then what would be the limit of this sequence?

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    Why do you think $l =1-\pi $? – 2017-01-21
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    No if you see the recurring formula the limit point at n step (say ), will always be greater than π . although in our calculation it comes out to be π and 1-π. – 2017-01-21

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If $x_n-\pi> 1$ then $$x_{n+1}-\pi =\sqrt {x_n-\pi}> \sqrt 1=1.$$ Since $x_1-\pi>1, $ we have $x_n-\pi>1$ for all $n$ by induction on $n.$

You have correctly shown that if the limit exists it is $\pi$ or $\pi +1.$ But since $x_n>\pi +1$ for all $n,$ the limit cannot be $\pi.$ The problem allows you to assume the limit exists (The allowable answers give you no other option.) So it must be $\pi+1.$

BTW since $y>1\implies y>\sqrt y>1,$ we have $x_n-\pi>1\implies x_n-\pi>x_{n+1}-\pi>1.$ So if $x_1>\pi +1$ then the sequence $(x_n-\pi)_n$ is decreasing and has a lower bound (namely, $1$) so it must have a limit.

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    How did u assume at the beginning that n term minus π is greater than 1 . This is a convergent sequence and hence its possible that the n^th term may be smaller than π. – 2017-01-21
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    I said IF $x_n-\pi>1$ THEN $x_{n+1}-\pi >1.$ I didn't assume anything. – 2017-01-21