Posted by **cara** on Wednesday, September 2, 2009 at 2:51pm.

how would I do this limit?(by the way

-> is an arrow(

lim x-> 0 (5x^3+8x^2)/ (3x^4 - 16x^2)

- math -
**MathMate**, Wednesday, September 2, 2009 at 3:22pm
When both the numerator and denominator evaluate to zero or infinity, the limit can be found by applying the d'hÃ´pital rule.

The rule states that if the limit of a division is indeterminate, the limit of the derivatives of the numerator and the denominator will equal the limit in question. The rule can be applied successively until either the denominator becomes defined and non-zero.

For the case in point,

let

numerator=f1(x)=(5x^3+8x^2)

denominator=f2(x)=(3x^4 - 16x^2)

Since f2(0)=0, we need to find the derivative

f21(x)=d(f1(x))/dx=12x^3-32x.

Again f21(0)=0, so we find the derivative

f22(x)=d²(f2(x))/dx²=36x²-32

f22(0)=-32

Now we have to evaluate

f12(x)=d²(f1(x))/dx²=30x+16

f12(0)=16

Therefore the limit x→0

(5x^3+8x^2)/ (3x^4 - 16x^2)

=f12(0)/f22(0)

=16/(-32)

=-(1/2)

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