Science Forums 2

[Don Blazys]

To: Qfwfq.

At x=1,

2x^6-2x^5+x^4+4x^3-12x^2+11x-4
-------------------------------------
3x^5+2x^4-5x^3+4x^2-12x+8

= 0/0, which is what we mathematicians refer to as an "indeterminate form". We can, however, show that in this particular case, the "indeterminate form" 0/0=1.25.

All we have to do is simply divide the derivative of the numerator by the derivative of the denominator and again let x=1. Thus we have:

12x^5-10x^4+4x^3+12x^2-24x+11
-------------------------------------
15x^4+8x^3-15x^2+8x-12

which at x=1 gives us 1.25.

This handy little technique is known as "L'Hopitals rule", and is quicker and more accurate than trying to "approach" x=1 as a "limit".

As for the equation:
a=e^(lna),
it is indeed an identity for all "natural numbers":
a={1,2,3...},
whereas the trivial:
a=a
is an identity for all "non-negative integers":
a={0,1,2...}.

Don.