what are the results? (1 Viewer)

[iluvmaths]

hello, I am new here
I am wondering what are the results of...
0 x infinity
infinity / infinity

one more thing, how to compare two complex number?
thx

 

[spice girl]

Originally posted by iluvmaths
I am wondering what are the results of...
0 x infinity
infinity / infinity

one more thing, how to compare two complex number?
thx

Hahaha...is the first question meant to confuse ppl?

Anyway, to compare two complex numbers, you compare either their moduli, or their arguments. You can't use inequality signs. Note that inequality signs only compare the moduli and arguments of real numbers anyway.

 

[Lazarus]

Originally posted by iluvmaths
hello, I am new here
I am wondering what are the results of...
0 x infinity
infinity / infinity

They are both indeterminate.

They only really arise when you're attempting to find limits... in the first case, you need to rearrange it so that either the zero or the infinity disappears, or you need to figure out which function 'grows' the fastest, and then take the limit of that function as the answer.

In the second case, you can either rearrange or use L'Hospital's Rule.

 

[Viator]

You shouldn't think of infinity as an actual number, its more like an idea or concept.

But I think that even an infinite amount of 0's would still be 0.

0 + 0 + 0 + ... + 0 = 0

 

[iluvmaths]

Re: Re: what are the results?

Originally posted by Lazarus
They are both undefined.

They only really arise when you're attempting to find limits... in the first case, you need to rearrange it so that either the zero or the infinity disappears, or you need to figure out which function 'grows' the fastest, and then take the limit of that function as the answer.

In the second case, you can either rearrange or use L'Hospital's Rule.

Lararus,
could you give me an example on either case?
And what is L'Hospital's Rule?

 

[Lazarus]

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⁺](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⁺, 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⁺, and so the limit is of the indeterminate oo/oo form and l'Hospital's Rule can be applied.

i.e.

lim[x -> 0⁺](x.lnx)
= lim[x -> 0⁺](ln x/(1/x))

apply l'Hospital's rule: differentiate the numerator and the denominator

= lim[x -> 0⁺]((1/x)/(-1/x²))
= lim[x -> 0⁺](-x)
= 0


In solving that question, another possible option would have been to write:

lim[x -> 0⁺](x.lnx) = lim[x -> 0⁺](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.

 

[spice girl]

Although you won't need L'hopital's rule in ext 2 maths...