What is the L'Hospital proof problem regarding limits at a point?

[pk1234]

Hey guys, first year university math student here. I need some help explaining the proof used in the scripts I'm studying from - just part of the proof to be more precise. English isn't my first language and I don't have much experience writing/rewriting down proofs and I don't know how to write those nice latex symbols, so sorry in advance if something doesn't make sense:


Presuming:
(1), a is element of R (|a| =/= +oo)
(2), f and g are real functions
(3), limit x->a_+ (f'(x) / g'(x)) exists (must be element of R, or +-oo)
(4), limit x->a_+ (f(x)) = limit x->a_+ (g(x)) = 0

then

limit x->a_+ (f(x))/(g(x)) = limit x->a_+ (f'(x))/(g'(x))



I think I understand most of the proof but there's something right at the start that I'm completely stuck at and still don't understand precisely enough:

Let L=limit x->a_+ (f'(x) / g'(x)).

There exists delta>0, such that for all x element of (a,a+delta), f and g are both defined on this interval,

- I think this can be proved easily from (4), correct? Also, |f| and |g| are both smaller than some Epsilon>0. The following however, I don't understand at all:

and both f' and g' have a finite (not = oo or -oo) derivation on this interval, and also g'=/=0.

Why is the derivation necessarily finite?


EDIT:

To explain where I see the problem a bit more precisely, let's say:

L=0
f(x)=0 for all x element R, and therefore f'(x)=0 for all x element R

Now, from limit x->a_+ (f'(x) / g'(x)) = 0 , it should be possible to somehow prove, that there exists a delta>0, such that for all x element (a,a+delta), g'(x) is finite and non zero. I really don't see it though, why can g'(x) not be +oo somewhere in that interval?

 

[Stephen Tashi]

Now, from limit x->a_+ (f'(x) / g'(x)) = 0 , it should be possible to somehow prove, that there exists a delta>0, such that for all x element (a,a+delta), g'(x) is finite and non zero. I really don't see it though, why can g'(x) not be +oo somewhere in that interval?

Pick \epsilon = 0.5 Since the above limit exists, there exists a \delta > 0 such that a < x < a + \delta implies | \frac{f'(x)}{g'(x) } - 0 | < 0.5

The statement | \frac{f'(x)}{g'(x)}| < 0.5 is not true unless the fraction \frac{f'(x)}{g'(x)} exists, i.e. is a specific number with an absolute value than can be compared to 0.5. When g'(x) is 0, the fraction doesn't exist. When g'(x) doesn't exist by virtue of being "equal" to \infty the fraction doesn't exist.

 

[pk1234]

Pick \epsilon = 0.5 Since the above limit exists, there exists a \delta > 0 such that a < x < a + \delta implies | \frac{f'(x)}{g'(x) } - 0 | < 0.5

The statement | \frac{f'(x)}{g'(x)}| < 0.5 is not true unless the fraction \frac{f'(x)}{g'(x)} exists, i.e. is a specific number with an absolute value than can be compared to 0.5. When g'(x) is 0, the fraction doesn't exist. When g'(x) doesn't exist by virtue of being "equal" to \infty the fraction doesn't exist.

Thanks I think I'm starting to see where the problem is -

When g'(x) doesn't exist by virtue of being "equal" to \infty the fraction doesn't exist.

Why does it not exist, if it's equal to +oo?

 

[Stephen Tashi]

Real valued functions exist at those real numbers where their values are real numbers. \infty is not a real number.

 

[pk1234]

Real valued functions exist at those real numbers where their values are real numbers. \infty is not a real number.

Why does g'(x) have to be a real valued function?

 

[Stephen Tashi]

The fraction | f&#039;(x)/g&#039;(x)| isn't comparable to the real number \delta by the relation "<" unless the fraction is a real number. The fraction isn't a real number unless it is the ratio of real numbers.

 

[pk1234]

Oooh. I thought that 0/oo = 0, and instead it is undefined?

 

[Stephen Tashi]

Oooh. I thought that 0/oo = 0, and instead it is undefined?

Yes, it's undefined. Don't confuse a ratio of numbers with limit of ratios.

 

[pk1234]

Thank you very much!