How do you find the limit of #((x^2 +2)/(x^3-1)# as x approaches infinity?

1 Answer
Gió
May 22, 2015

Imagine that we use as #oo# a values very big such as: #1,000,000#
Your function, basically, becomes: #(1,000,000)^2/(1,000,000)^3# the other two numbers #+2# and #-1# are irrelevant compared to #1,000,000# so you can forget them.

Now you have:

#cancel((1,000,000)^2)/(1,000,000)^cancel(3)=1/(1,000,000)~=0#

So you can write:

#lim_(x->oo)(x^2+2)/(x^3-1)=0#

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