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# L'Hôpital's rule introduction

When you are solving a limit, and get 0/0 or ∞/∞, L'Hôpital's rule is the tool you need. Created by Sal Khan.

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• Isn't infinity over infinity just equal to 1, or negative infinity over infinity equal to -1? My brain thinks of it like infinity is a value and dividing it by itself gets 1.
• But infinity isn't a number or a value.

Consider what x^2 / x is as x goes to infinity: you get ∞/∞. However, x^2 / x can be simplified to x, and as x goes to infinity you get ∞. So, ∞/∞ is equal to ∞?

Now consider x / x^2 as x goes to infinity: you get ∞/∞ again. But x / x^2 can also be simplified to 1/x, and as x goes to infinity you get 0. So now ∞/∞ is equal to 0?

∞/∞ is very much undefined, because ∞ is not a value or a number.
• even after applying the l'hopital's rule, if it remains in 0/0 or other indefined form then?
• If it remains 0/0 or ∞/∞ then you can repeat l'Hopital's rule.
However, sometimes l"Hopital's will never produce determinate form, so you have to solve the limit by some other means.
• Who is L'Hopital
• At , Sal says there are indeterminate forms like 0/0 and infinity/infinity. Are there any other types?
Also, what makes another form indeterminate, i.e., what is the rationale?

Thanks!
• A form is indeterminate if we can't tell what the limit is just by looking at the form. For example, a form that looks like 0/0 or infinity/infinity could end up having limit 0, infinity, -100, 0.5, undefined, or any real number.
Examples of 0/0 cases:
1. limit_{x->0} (sin x)/x = 1
2. limit_{x->0} (1- cos x)/x = 0
3. limit_{x->0} x/(x^2) = limit_{x->0} 1/x which does not exist.
As you can see, just having the form 0/0 doesn't tell us anything about the value of the limit.

Some other indeterminate forms are infinity - infinity, 1^infinity, 0*infinity.
But note that things like infinity+infinity and 0^infinity are NOT indeterminate forms.
• Okay, so I know how about L'Hopital's rule and what it is, maybe a bit about how to use it. But why does it work? Can anyone prove L'Hopital's rule?
• If lim f(x) is defined and lim g(x) is defined, we wouldn't need L'Hopital's rule to find lim f(x)/g(x), but would it still apply?
• L'Hôpital's rule can only be applied in the case where direct substitution yields an indeterminate form, meaning 0/0 or ±∞/±∞. So if f and g are defined, L'Hôpital would be applicable only if the value of both f and g is 0.

Think about the limit of (x+1)/(x+2) as x approaches 0. Direct substitution tells us that the answer is 1/2. However, if we tried to apply L'Hôpital's rule, we would get 1 as our answer, which would be incorrect.
• Why does this work? It is great that it works, but why?
• What do you do if you get infinity times zero when plugging in as a test for l'hopital?
• That is not a correct form for l'Hôpital's rule, so it is still indeterminate. You need to convert it to something that is a l'Hôpital's form.
Specifically,
If g(x) → 0
And f(x)→ ∞
Then: g(x) f(x) is the form you mentioned.
but f(x) = 1 / [ 1/f(x) ]
And 1/f(x) is a 0 form.
Thus,
g(x) f(x) = g(x) / [1/f(x)]
and g(x) / [1 / f(x) ] is a 0/0 form and subject to l'Hôpital's rule
• Does this only apply to fractions