Question

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

Answer 1

`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.