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From: Candide Voltaire on 16 Jan 2010 05:18
consider the following trivial example:
lim for x-->0 (x/x)
applying de L'Hopitals rule gives 1
However 0/0 can be any number not just 1
How then can I be sure when I use de L'Hopital for complex expressions
it wil not hide solutions

candide
From: Tonico on 16 Jan 2010 05:51
On Jan 16, 12:18 pm, Candide Voltaire <candideguev...(a)gmail.com>
wrote:
> consider the following trivial example:
> lim for x-->0 (x/x)
> applying de L'Hopitals rule gives 1
> However 0/0 can be any number not just 1


0/0 is not a number and thus can't be any of them. L'Hospital's rule
is a theorem and you can
use it whenever that theorem's conditions are fulfilled. Its being a
theorem makes it sure that
what you get is correct.

Tonio


> How then can I be sure when I use de L'Hopital for complex expressions
> it wil not hide solutions
>
> candide

From: Aielyn on 16 Jan 2010 05:52
On Jan 16, 8:18 pm, Candide Voltaire <candideguev...(a)gmail.com> wrote:
> lim for x-->0 (x/x)
> applying de L'Hopitals rule gives 1
> However 0/0 can be any number not just 1
> How then can I be sure when I use de L'Hopital for complex expressions
> it wil not hide solutions

You aren't quite understanding the concept of limits.

Consider x/x. When x=1, x/x=1. When x=1/4, x/x=1. When x=10^(-10000),
x/x=1. No matter how close to 0 you get without reaching 0, x/x=1.
Therefore, the limit as x -> 0 of x/x is 1.

L'Hopital's rule applies to such limits, when both numerator and
denominator tend to zero (or both tend to infinity). It is not
equivalent to evaluating the values *at* x=0. For instance, the dirac
delta function is zero for any value of x except for x=0. The limit as
x->0 is zero, but it is infinite at x=0.
From: Jon Slaughter on 16 Jan 2010 06:18
Aielyn wrote:
> On Jan 16, 8:18 pm, Candide Voltaire <candideguev...(a)gmail.com> wrote:
>> lim for x-->0 (x/x)
>> applying de L'Hopitals rule gives 1
>> However 0/0 can be any number not just 1
>> How then can I be sure when I use de L'Hopital for complex
>> expressions it wil not hide solutions
>
> You aren't quite understanding the concept of limits.
>
> Consider x/x. When x=1, x/x=1. When x=1/4, x/x=1. When x=10^(-10000),
> x/x=1. No matter how close to 0 you get without reaching 0, x/x=1.
> Therefore, the limit as x -> 0 of x/x is 1.
>
> L'Hopital's rule applies to such limits, when both numerator and
> denominator tend to zero (or both tend to infinity). It is not
> equivalent to evaluating the values *at* x=0. For instance, the dirac
> delta function is zero for any value of x except for x=0. The limit as
> x->0 is zero, but it is infinite at x=0.

The trivial example may be what is confusing him

take f(x) = (x^2 - 1)/(x + 1)

Note that this is not the same as g(x) = x-1

They are two different functions.

g(x) is defined at x = -1 while f(x) is not. They are idential for all other
x values BUT ARE DISTINCT AND DIFFERENT FUNCTIONS.

f(-1) is undefined but g(-1) = -2.

In some cases we treat g(x) and f(x) interchangably(such as when we simplify
expressions as we are taught to do in precal and algebra class... but we
must keep in mind there is a distinct difference between them when it
matters(it most cases it does not matter).

lim x->-1 f(x) = lim x->-1 g(x) but f(-1) != g(-1) (notice that this is the
basic requirements for continuity and says f(x) is not continuous but g(x).

0/0 is not a function as it has no meaning. 0/0 is simply an undefined
quantity.

Take f(x) = |x|/x.

f(0) = 0/0 which is undefined...

but this entirely different than lim x->0 f(x) which is still undefined in
this case simply beacuse the left and right limits are not the same.


f(x) = x/x does not exist for x = 0. f(x) = 1 for all x != 0.

i.e.,

f(x) = 1 for all x!= 0 is not the same as
f(x) = 1.

they are different, they may be closely related but have a significant
different. You need to realize that when you "Simplify" you are throwing
away information. Simplification is what is "hiding" your solutions.
(basically the point is you can't blindly simplify something then expect it
to still be exactly the same as before)

Similarly 9/2 is different than 4.5. One is a faction and the other a
decimal. In this case the difference is only visual and generally we treat
the two as identical(because they are identical numerically).

For f(x) = x/x and g(x) = 1, they are identical for all x != 0. At x = 0
they are not identical. We generally just treat f and g as equivalent but as
you can see, blindly following that can lead you astray.

From: porky_pig_jr on 16 Jan 2010 16:19
On Jan 16, 5:18 am, Candide Voltaire <candideguev...(a)gmail.com> wrote:
> consider the following trivial example:
> lim for x-->0 (x/x)
> applying de L'Hopitals rule gives 1
> However 0/0 can be any number not just 1
> How then can I be sure when I use de L'Hopital for complex expressions
> it wil not hide solutions
>
> candide

Understand what x --> 0 means first.
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