Question

Question #31ec4

Answer 1

`lim_(x->oo)(2x^2 - 1/4x^3 - 5x - 1) = -oo`

Explanation:

As we increase the value of `x`, the term which will increase the fastest is the one with the highest degree, which is `-1/4x^3` in this case.

In fact, as you take any polynomial to infinity, the only term that will matter is the one with the biggest degree, since it will dominate all of the others as far as rate of growth.

Therefore, we can say that:

`lim_(x->oo)(2x^2 - 1/4x^3 - 5x - 1)`
`= lim_(x->oo)(cancel(2x^2) - 1/4x^3 - cancel(5x) - cancel(1))`
`= lim_(x->oo)(-1/4x^3)`

And as we plug bigger and bigger values of `x` into `-1/4x^3`, it will grow faster and faster towards `-oo`, since `x^3` grows towards positive infinity, and multiplying that by a negative number makes it grow towards negative infinity.

Final Answer

Answer 2

`0`

Explanation:

`"divide terms on numerator/denominator by the "`
`"highest power of x that is "x^3`

`rArr((2x^2)/x^3-1/x^3)/((4x^3)/x^3-(5x)/x^3-1/x^3)=(2/x-1/x^3)/(4-5/x^2-1/x^3)`

`rArrlim_(xtooo)(2x^2-1)/(4x^3-5x-1)`

`=lim_(xtooo)(2/x-1/x^3)/(4-5/x^2-1/x^3)`

`=0/(4-0-0)=0`

Answer 3

`0`

Explanation:

For `(2x^2-1)/(4x^3-5x-1)`

Since `x^2 and x^3` are changing more rapidly than `x`, we can omit `x` and the constants from the expression. Then we have:

`(2x^2)/(4x^3) = (x^2)/(2x^3)`

Dividing by `x^2`:

`1/(2x)`

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