What is the limit as #x -> ∞# of #(x^2 + 2) / (x - 1)#?

1 Answer
Feb 3, 2016

This function diverges; the limit is 'equal' to #oo#.

Explanation:

Divide both the numerator and the denominator by #x#, that makes it easier.
#lim_(n to oo) (x^2/x+2/x)/(x/x-1/x)=lim_(n to oo)(x+2/x)/(1-1/x)#

We know that #lim_(n to oo) 1/x=0#, so
#lim_(n to oo)(x+2/x)/(1-1/x)=lim_(n to oo)(x+0)/(1-0)=lim_(n to oo)(x)/(1)=lim_(n to oo)x=oo#