Limit as x approaches infinity and negative infinity?

For infinity limits, you would divide every term by the highest degree term, so for this case x^2 is the highest degree term, so after dividing your function would be...

1/x + 32/(x^2)

Now just substitute in infinity into the function, and the answer should be zero.

If you don`t get it, basically any number divided by infinity will always equal zero, since any number divided by zero equals infinity.

You would also get the same answer for as the function approaches negative infinity.

limit as x approaches +infinity = 0

limit as x approaches -infinity = 0

The reasoning: Divide numerator and denominator by by x that is multiply by (1/x) / (1/x) since this is equal to one we don't "change" the equation at all, but we get the new fraction:

(x/x + 32/x) / x^2 / x

Simplify to get (1 + 32/x) / x

We know that 1/x^a where a > 0 yields the value 0 as x approaches infinity, therefore we can say that 32/x = 0 as x approaches infinity, we now have 1 / x, and once again use this identity to see the limit approaches 0.

The negative infinity evaluation is the same.