Show that ( lim x n ) 2 = a and thus exhibit the existence of a positive square root of a. (because we took x 1 &gt; 0)

Question

Show that   (  lim      x    n        )    2    =  a and thus exhibit the existence of a positive square root of   a. (because we took       x    1    &amp;gt;  0)

Colin Moran · Accepted Answer

You&#039;ve done the hardest part. Let             lim              n        →        ∞                    x      n        =    x  . Rearranging the recursion formula gives  2  ⋅      x    n    ⋅      x          n      +      1        =      x    n    2    +  aTaking limits and noting that             lim              n        →        ∞                    x      n        =          lim              n        →        ∞                    x              n        +        1              =    x  , we have  2      x    2    =      x    2    +  aso       x    2    =  a.

Reginald Delacruz · Accepted Answer

If       x    n    →  x, then      x          n      +      1        =                    x        n        2            +      a              2              x        n              →  aas well. But                    x        n        2            +      a              2              x        n              →                    x        2            +      a              2      x        .Thus the limit x satisfies the equation  x  =                    x        2            +      a              2      x          ⟺    2      x    2    =      x    2    +  a    ⟺        x    2    =  a    ⟺    x  =  ±      a    .But as the sequence has terms terms it can not converge to a negative number. Thus       x    n    →      a  .

Show that ( <mo movablelimits="true" form="prefix">lim x n </msub>

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