Question

The intuition behind L'hopital's rule and the conditions which make it applicable?

Limx->a(f(x)/g(x))= Limx-->a(f'(x)/g'(x))
Provided that f(a)=0 and g(a)=0 and lim x->a f'(x)/g'(x) exist. Does the second condition imply that both functions, f(x) and g(x) are continuous and differentiable on some interval containing (a,f(a))? If both or either functions derivatives do not exist as the variable point x moves towards the fixed point (a,f(a)) then the limit of the quotient f'(x)/q'(x) does not exist. Another condition which I talso think is valid, g(x) has no critical points in the region contanining x=a.

Limx->a (f (x) − f (a)/x-a /g(x) − g(a)/x-a)=f'(a)/g'(a)

How do we proof L'Hoptial's rule in the case of infinity/infinity
The algebraic manipulation we did above will not work because as x approaches a the functions, f(x) and g(x), and f'(a) seem to me quite illogical because infinity is not a real number, yes the function might be heading in some definite direction but not in terms of output values

Answer 1

I'll try to answer some of your questions here.

Explanation:

`lim_(xrarra)(f'(x))/(g'(x))` is only defined if `f'(x)` and `g'(x)` exist in some interval containing `a`.

And that implies that `f` and `g` are continuous in some interval containing `a`.

There are many cases in which l'Hospital's rule does not work. You have described (at least) some of them.

If the limit of the quotient of derivatives exists or fails to exist because the quotient increases or decreases without bound, then we can use l'Hospital.
That is: if `lim_(xrarra)(f'(x))/(g'(x))` is some real number or `lim_(xrarra)(f'(x))/(g'(x)) = oo` or `lim_(xrarra)(f'(x))/(g'(x)) = -oo`, then we can use l'Hospital.

(I may be mistaken about the infinite case.)

It is quite possible that `g` has a critical point of type `g'=0` at `a`. If `f'(a)` is also `0`, we try to use l'Hospital again. If `lim_(xrarra)(f'(x))/(g'(x)) = oo` or `= -oo`, the l'Hospital applies and `lim_(xrarra)(f(x))/(g(x)) = oo` or `= -oo`

There is no need to treat infinity as a number . Writing

`lim_(xrarra)f(x) = oo` is a short way of saying

`lim_(xrarra)f(x)` does not exist because as `xrarra`, the values of `f(x)` increase without bound.

More precisely: `lim_(xrarra)f(x) = oo`

if and only if for any `M`, there is a positive `delta` such that for all `x`, if `0 < abs(x-a) < delta`, then `f(x) > M`.

That's a lot to write, so it is abbreviated by `lim_(xrarra)f(x) = oo`.

Rather than try to reproduce a proof for the `+-oo/oo` case, here, I refer you to the Wikipedia page https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Or other pages you may find if you search for l'hopital proof (or something similar).