What is # lim_(->-oo) f(x) = x^2/(x^2-7) -(2x^2)/(x^3-5)#?

1 Answer
May 18, 2018

#lim_(x-> -infty) f(x) = 1#

Explanation:

We'll take each of the two rational terms and divide through by their highest power of #x#. This gives:

#f(x) = x^2 / (x^2 - 7) - (2x^2)/(x^3-5) = 1 / (1 - 7/x^2) - (2/x)/(1 - 5/x^3#

Now recall the fact that #lim_(x-> pm infty) a / x^n = 0# if #a# is a real number and #n >= 1#. This tells us that #7/x^2#, #2/x# and #5/x^3# will all approach #0# as #x -> -infty#. Thus, we can directly replace these terms with zero which yields our limit.

#lim_(x-> -infty) 1 / (1 - 7/x^2) - (2/x)/(1 - 5/x^3) = 1 / (1-0) - 0 / (1-0) = 1#.

This is our final answer.