Question #12c6a

1 Answer
Archish R.
Jan 9, 2018

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

Explanation:

#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#

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