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Evaluate limits. lim x-->infinity sin(x-1/2+x^2) lim x-->1+ Sqrt(x+1) -1 lim...
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High School Teacher
(1) since `lim_(x->oo) x-1/2+x^2 = oo`
and `sin x` does not have a limit as `x->oo` because it does not approach any number but varies between -1 and 1, then
`lim_(x->oo) sin(x-1/2+x^2)` does not exist
(2) `lim_(x->-1^+) sqrt(x+1)-1=lim_(x->-1^+) sqrt(x+1) - lim_(x->-1^+)1`
from the right side `lim_(x->-1+) sqrt(x+1) = 0` , and `lim_(x->-1^+) = 1`
So `lim_(x->-1^+) sqrt(x+1)-1 = 0 - 1 = -1`
(3) Really the same issue as (1) `lim_(x->oo) x^2+1/x+1 = oo` so the limit of sin does not exist for the same reason as in (1).
Posted by beckden on September 2, 2012 at 9:38 PM (Answer #1)
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