how would I do this limit?(by the way
-> is an arrow(
lim x-> 0 (5x^3+8x^2)/ (3x^4 - 16x^2)
When both the numerator and denominator evaluate to zero or infinity, the limit can be found by applying the d'hôpital 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)