# L'Hospital proof problem

by pk1234
Tags: lhospital, proof
 P: 11 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?
P: 3,296
 Quote by pk1234 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.
P: 11
 Quote by Stephen Tashi 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?

 Sci Advisor P: 3,296 L'Hospital proof problem Real valued functions exist at those real numbers where their values are real numbers. $\infty$ is not a real number.
P: 11
 Quote by Stephen Tashi 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?
 Sci Advisor P: 3,296 The fraction $| f'(x)/g'(x)|$ 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.
 P: 11 Oooh. I thought that 0/oo = 0, and instead it is undefined?