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submitted 10 months ago by siriusmart to c/dailymaths

5 comments fedilink hide all child comments

Prove L'Hopital's rule, nothing fancy.

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[–] [email protected] 7 points 10 months ago (1 children)

I have a truly marvelous demonstration of this proposition which this comment is too narrow to contain

[–] [email protected] 5 points 10 months ago

CuriousReFermat strikes again

[–] zkfcfbzr 5 points 10 months ago* (last edited 10 months ago) (1 children)

Here's a very shitty wishy-washy proof that takes a few liberties with what you can do with limits:

f'(x) = lim x → h of (f(x+h) - f(x)) / h and g'(x) = lim h → 0 of (g(x+h) - g(x)) / h.

So f'(x)/g'(x) = (lim h → 0 of (f(x+h) - f(x)) / h) / (lim h → 0 of (g(x+h) - g(x)) / h)

= lim h → 0 of (f(x+h) - f(x)) / (g(x+h) - g(x))

So lim x → a of f'(x)/g'(x) = lim x → a lim h → 0 of (f(x+h) - f(x)) / (g(x+h) - g(x))

Plug in x = a and the - f(x) and - g(x) terms disappear, since we're given f(a) = g(a) = 0. Then un-plug-in x = a to keep the rest of the limit.

lim x → a of f'(x) / g'(x) = lim x → a lim h → 0 of f(x+h) / g(x+h)

Plug in h = 0 limit

lim x → a of f'(x) / g'(x) = lim x → a of f(x) / g(x), QED ¯\_(ツ)_/¯

[–] siriusmart 4 points 10 months ago

thats the wrong way round, but yeah its the standard way of proving it

[–] siriusmart 4 points 10 months ago* (last edited 10 months ago)

I know a standard way of proving it, and a much shorter way of proving it, don't know if you know another other "non standard" proofs.

hint:

spoiler

linear approximation

spoiler