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

  • Thread starter pk1234
  • Start date
  • Tags
    Proof
In summary, the proof used in the scripts I'm studying requires that there exists a delta>0 such that for all x in (a,a+delta), g'(x) is finite and nonzero. However, I don't understand why g'(x) can't be +oo somewhere in that interval.
  • #1
pk1234
11
0
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?
 
Last edited:
Physics news on Phys.org
  • #2
pk1234 said:
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 [itex] \epsilon = 0.5 [/itex] Since the above limit exists, there exists a [itex] \delta > 0 [/itex] such that [itex] a < x < a + \delta [/itex] implies [itex] | \frac{f'(x)}{g'(x) } - 0 | < 0.5 [/itex]

The statement [itex] | \frac{f'(x)}{g'(x)}| < 0.5 [/itex] is not true unless the fraction [itex] \frac{f'(x)}{g'(x)} [/itex] exists, i.e. is a specific number with an absolute value than can be compared to 0.5. When [itex] g'(x) [/itex] is 0, the fraction doesn't exist. When [itex] g'(x) [/itex] doesn't exist by virtue of being "equal" to [itex] \infty [/itex] the fraction doesn't exist.
 
  • #3
Stephen Tashi said:
Pick [itex] \epsilon = 0.5 [/itex] Since the above limit exists, there exists a [itex] \delta > 0 [/itex] such that [itex] a < x < a + \delta [/itex] implies [itex] | \frac{f'(x)}{g'(x) } - 0 | < 0.5 [/itex]

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

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

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

Why does it not exist, if it's equal to +oo?
 
  • #4
Real valued functions exist at those real numbers where their values are real numbers. [itex] \infty [/itex] is not a real number.
 
  • #5
Stephen Tashi said:
Real valued functions exist at those real numbers where their values are real numbers. [itex] \infty [/itex] is not a real number.

Why does g'(x) have to be a real valued function?
 
  • #6
The fraction [itex] | f'(x)/g'(x)| [/itex] isn't comparable to the real number [itex] \delta [/itex] 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.
 
Last edited:
  • #7
Oooh. I thought that 0/oo = 0, and instead it is undefined?
 
  • #8
pk1234 said:
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.
 
  • #9
Thank you very much!
 

What is the L'Hospital proof problem?

The L'Hospital proof problem, also known as the L'Hospital's rule or the L'Hospital's rule for indeterminate forms, is a mathematical theorem that helps to evaluate limits where the numerator and denominator both approach zero or infinity, resulting in an indeterminate form. It provides a way to simplify the evaluation of these limits by using derivatives.

What is the history behind the L'Hospital proof problem?

The L'Hospital proof problem was first discovered by French mathematician Guillaume de l'Hospital in the late 17th century. However, it was actually first discovered by Swiss mathematician Johann Bernoulli, who communicated it to L'Hospital. L'Hospital then published it in his book "Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes" in 1696.

What are the conditions for applying L'Hospital's rule?

In order to apply L'Hospital's rule, the limit must be in an indeterminate form, meaning that both the numerator and denominator approach zero or infinity. Additionally, the limit must be of the form 0/0 or ∞/∞. If these conditions are met, then L'Hospital's rule can be applied to simplify the evaluation of the limit.

What is the process for applying L'Hospital's rule?

To apply L'Hospital's rule, first, take the derivative of the numerator and denominator separately. Then, evaluate the limit of the resulting function. If the limit still results in an indeterminate form, the process can be repeated until an answer is obtained. It is important to note that L'Hospital's rule only works for certain types of indeterminate forms and may not always give the correct answer.

What are some examples of using L'Hospital's rule?

One example of using L'Hospital's rule is when evaluating the limit limx→0 (sin x)/x. This limit is in the form 0/0, so L'Hospital's rule can be applied. The derivative of sin x is cos x, and the derivative of x is 1. Therefore, the limit becomes limx→0 cos x = 1. Another example is when evaluating the limit limx→∞ (x² + 3x + 1)/(2x² + 5). This limit is in the form ∞/∞, so L'Hospital's rule can be applied. Taking the derivatives results in limx→∞ (2x + 3)/(4x) = 1/2.

Similar threads

  • Calculus
Replies
9
Views
2K
Replies
5
Views
271
Replies
9
Views
891
Replies
4
Views
641
Replies
1
Views
831
Replies
1
Views
1K
  • Calculus
Replies
5
Views
1K
Replies
2
Views
1K
Replies
14
Views
1K
Back
Top