Question #12c6a

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Answer 1 · 2018-01-09T09:10:04

$limx-> oo (sqrtx+1-1/x)=oo$

$limx-> oo (sqrtx+1-1/x)$

$limx-> oo ((xsqrtx+x-1)/x)$

this is $oo/oo$ , so we can use l'Hôpital's Rule

$limx-> oo ((d/dx(xsqrtx+x-1))/(d/dxx))$

$=limx-> oo(((3/2sqrtx+1))/1)$

When we plug in infinity for x,

$=3/2sqrtoo+1=oo$