can some one help me with this problem please limit x tending to infinity 2x^3/-3x^2-4x-4?

Question

limit x tending to infinity 2x^3/(-3x^2-4x-4)

Answer 1

devide by x^2 to both numerator and denomenator

we get -2x/(3 -4/x-4/x^2)
as x->infinite numerator -> -inf and denominator tends to 3 so limit -> - infinite

Answer 2

2x^3 / (-3x^2-4x-4) lim x tending to infiniti

to solve such problem always divide the numerator and denominator by highest power of x in the equation, here its x^3.

dividing numerator and denominator by x^3 we get;

-2 / ( (3/x) + (4/x^2) + (4/x^3) )

here as x tends to infinity 3/x tends to 0, 4/x^2 tends to 0 , and 4/x^3 tends to 0.

so denominater tends to 0.

thus answer is -2/0

that is minus infiniti

Answer 3

divide numerator and denominator by x^3 as it is a limit, not actually infinity.
it becomes
2/(-3(x^-1)-4(x^-2)-4(x^-3))
denominator, on applying limits tends to 0 through negative values.
so the fraction tends to -infinity (right handed limit)

Answer 4

The x^3 in the numerator trumps everything else.

For a sense of how this is so, enter a large number for x in the equation, like 10000. The absolute value of the answer will be HUGE.

With a HUGE x value, the answer will be negative HUGE.

So the answer is negative infinity!

Answer 5

The answer is minus infinity,because if you use l`Hopital you get:
6x^2/(-6x-4)
12x/-6=-2x
and limit(-2x),where x=infinity is minus infinity

the answer I`m sure is a good one

Answer 6

x * 2/ [( -3 -4/x - (4/x^2)]

4/x->0
4/x^2 -> 0


=> lim (x-> inf) from ( - 2/3 * X) ....which is (-infinity)

Answer 7

Abe khamakha bheja kyun paka rela he ?

Answer 8

huh sorry no idea