How do you evaluate the limit #(x^2-x-6)/(x+2)# as x approaches #oo#?

1 Answer
georgef
Aug 21, 2016

The limit of a sum is the sum of the limits, provided the individual limits exist. Similarly with product, quotient, etc.

Explanation:

Let's manipulate for the expression to a more convenient form #(x^2-x-6)/(x+2)=((x^2-x-6)/x)/((x+2)/x)=((x^2/x-x/x-6/x))/((x/x+2/x))=(x-1-6/x)/(1+2/x)#

Then #lim_(xrarroo) ((x^2-x-6)/(x+2)) = lim_(xrarroo) ((x-1-6/x)/(1+2/x)) = (oo -1 -0)/(1+0)=oo#

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