# Find `lim_(x->-oo) x^2 - x^4`

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The limit `lim_(x->-oo) x^2 - x^4` has to be determined.

`lim_(x->-oo) x^2 - x^4`

As x tends to `-oo`,  ` x^4` decreases at a rate faster than `x^2` as it is raised to a higher power. The result is the expression `x^2 - x^4` tending to `-oo` as `x` tends to `-oo` .

The value of `lim_(x->-oo) x^2 - x^4 = -oo`

Sources:

beckden | High School Teacher | (Level 1) Educator

Posted on

`lim_(x->-oo)(x^2-x^4)=lim_(x->-oo)x^2(1-x^2) `

We now have `lim_(x->-oo)x^2*lim_(x->-oo)1-x^4`

`lim_(x->-oo)x^2 = oo and lim_(x->-oo)(1-x^4)=-oo`

So `lim_(x->-oo)(x^2-x^4) = -oo`

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`lim_(x->-oo )x^2 - x^4`

Factor out the x with the highest power.

`lim_(x->-oo ) x^4 (1/x^2 - 1)`

Then, take the limit of each factor separately.

`lim_(x->-oo) x^4 * lim_(x->-oo)(1/x^2 - 1)`

For the first factor, consider the property of limit which is `lim_(x->oo)cx^n = oo` .

But, take note the negative sign of infinity symbol in the above problem. A negative large number when raised with an even power results to a positive large number. Then,

`lim_(x->-oo) x^4 = oo`

For the second factor, use the property `lim_(x->+-oo)1/x^n = 0` .

`lim_(x->-oo) (1/x^2-1)= 0 - 1 = -1`

Then, combining the limit of each factor results to:

`lim_(x->-oo) (x^4 )(1/x^2-1) = oo(-1) = -oo`

The product is negative as per rules of multiplication for unlike signs. A positive large number when multiplied with a negative number, the product is negative.

Hence, `lim_(x->-oo) (x^2 - x^4) = -oo`