Originally posted by Viator
But I think that even an infinite amount of 0's would still be 0.
0 + 0 + 0 + ... + 0 = 0
Even so, mathematically, the terms 0/0, infinity/infinity, 0/infinity, infinity/0, 0 * infinity etc etc are all indeterminate.
iluvmaths:
I'll use oo to mean infinity.
l'Hospital's Rule:
Suppose f and g are differentiable and g'(x) ≠ 0 near a (except possibly at a). Suppose that:
lim[x -> a](f(x)) = 0 and lim[x -> a](g(x)) = 0
OR
lim[x -> a](f(x)) = oo and lim[x -> a](g(x)) = oo
(In other words, the composite limit of f(x)/g(x) is an indeterminate form of type 0/0 or oo/oo.) Then
lim[x -> a](f(x)/g(x)) = lim[x -> a](f'(x)/g'(x))
if the limit on the right side exists (or is oo or -oo).
example:
lim[x -> 0<sup>+</sup>](x.ln(x))
To begin with, it isn't clear what the value of this limit will be (if it even exists). There is a struggle between x and ln x. If x wins, the answer will be 0; if ln x wins, the answer will be oo (or -oo). Or there may be a compromise where the answer is a finite nonzero number. This kind of limit is called an indeterminate form of type 0*oo. We can deal with it by writing the product as a quotient, which would convert the limit to an indeterminate form of type 0/0 or oo/oo so that we can use l'Hospital's Rule.
This limit is indeterminate because, as x -> 0<sup>+</sup>, the first factor (x) approaches 0 while the second factor (ln x) approaches -oo. Writing x = 1/(1/x), we have 1/x -> oo as x -> 0<sup>+</sup>, and so the limit is of the indeterminate oo/oo form and l'Hospital's Rule can be applied.
i.e.
lim[x -> 0<sup>+</sup>](x.lnx)
= lim[x -> 0<sup>+</sup>](ln x/(1/x))
apply l'Hospital's rule: differentiate the numerator and the denominator
= lim[x -> 0<sup>+</sup>]((1/x)/(-1/x<sup>2</sup>))
= lim[x -> 0<sup>+</sup>](-x)
= 0
In solving that question, another possible option would have been to write:
lim[x -> 0<sup>+</sup>](x.lnx) = lim[x -> 0<sup>+</sup>](x/(1/ln x))
This gives an indeterminate form of the type 0/0, but if we apply l'Hospital's Rule we get a more complicated expression than the one we started with. In general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit.