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How do you find #lim (5x+6)/(x^2-4)# as #x->oo#?

1 Answer

Jan 26, 2017

The answer is #=0^-# and #0^+#

Explanation:

#lim_(x->-oo)(5x+6)/(x^2-4)=lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))#

#lim_(x->-oo)6/x=0#

#lim_(x->-oo)4/x^2=0#

#lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->-oo)5/x=0^-#

#lim_(x->+oo)(5x+6)/(x^2-4)=lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))#

#lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->+oo)5/x=0^+#

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