How do you find #lim (6x^3-4)/(2x+5)# as #x->oo#?

1 Answer
Gió
Apr 5, 2017

You should get #oo#

Explanation:

Consider the numerator first: even if you subtract #4# as #x# increases it will become very big and rapidly.
The denominator as well BUT not as fast as the numerator!
So the numerator will "win"and in the limit the function will tend to #oo#.

We can also write collecting #x#:
#lim_(x->oo)(cancel(x)(6x^2-4/x))/(cancel(x)(2+5/x))#
as #x->oo#
#4/x->0# and #5/x->0# giving:
#lim_(x->oo)(cancel(x)(6x^2-4/x))/(cancel(x)(2+5/x))=oo#

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