Question

Question #8da62

Answer 1

First, divide both the numerator and denominator by `x^2`.

`lim_(x->-oo)(-2x^5 + x^4 - 3) / (3x^2-7)`

`lim_(x->-oo)(-2x^3 + x^2 - 3/x^2)/(3-7/x^2)`

The `3/x^2` and `7/x^2` terms will go to `0` as `x ->oo`, so we can eliminate them from the expression.

`lim_(x->-oo)(-2x^3 + x^2)/3`

Now, we can prove that this limit equals `oo`.

Since `x^3` is an odd power, it will tend towards `-oo` as `x -> -oo`.
Therefore, multiplying `x^3` by a negative number will make it tend towards `+oo`.

Therefore, `-2x^3` will tend towards `+oo` as `x->-oo`.

Since `x^2` is an even power, it will tend towards `+oo` as `x -> -oo`

Since BOTH of the top terms will tend towards `+oo`, and the bottom term remains constant, the entire fraction will tend towards `+oo`. In other words:

`lim_(x->-oo)(-2x^3+x^2)/3 = +oo`

`therefore lim_(x->-oo)(-2x^5+x^4-3)/(3x^2-7) = +oo`