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What is domain of continuaty of y=lim ((x^2+1)^n+x-2)/(2(x^2+1)^n+x^2+2) , n go to...

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drifterkay | Student, Undergraduate | eNoter

Posted December 28, 2012 at 10:48 AM via web

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What is domain of continuaty of y=lim ((x^2+1)^n+x-2)/(2(x^2+1)^n+x^2+2) , n go to infinte?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted January 3, 2013 at 8:06 AM (Answer #1)

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You need to notice that`x^2 + 1 > 0`  for `x in R`   and `x^2 + 1 = 1`  if `x = 0` , hence, you may evaluate `f(0)`  such that:

`f(0) = lim_(n->oo) (1^n - 2)/(2*1^n + 2) = (1 - 2)/4 = -1/4`

You need to evaluate `f(x)`  such that:

`f(x) = lim_(n->oo) ((x^2+1)^n+x-2)/(2(x^2+1)^n+x^2+2)`

You need to force the factor `(x^2+1)^n`  such that:

`f(x) = lim_(n->oo) ((x^2+1)^n(1 + (x - 2)/((x^2+1)^n)))/((x^2+1)^n(2 + (x^2+2)/((x^2+1)^n)))`

Reducing duplicate factors yields:

`f(x) = lim_(n->oo) (1 + (x - 2)/((x^2+1)^n))/(2 +(x^2+2)/((x^2+1)^n))`

Since `lim_(n->oo)(x - 2)/((x^2+1)^n) = 0`  and `lim_(n->oo) (x^2+2)/((x^2+1)^n) = 0`  yields:

`f(x) = (1 + 0)/(2 + 0) = 1/2`

Hence, evaluating `f(x)`  yields `f(x) = {(1/2, x in R-{0}),(-1/4, x = 0):}, ` hence, the function is continuous over `R - {0}` .

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