How do you determine the limit of #(2x^2-4x) / (x+1)# as x approaches positive infinity?

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Answer 1 · 2016-07-02T18:06:39

# = oo#

from inspection you can see that the dominant term is the #x^2# term in the numerator so you can probably already guess where this is going

Doing it in a more "mathsy" way:

#lim_{x to oo} (2x^2-4x) / (x+1)#

#=lim_{x to oo} 2x*(x-2) / (x+1)#

#=2 lim_{x to oo} x* lim_{x to oo} (1-2/x) / (1+1/x)#

#= 2 lim_{x to oo} x#

# = oo#