Prove L’Hopital’s rule, nothing fancy.

  • CuriousRefugee
    6·
    7 months ago

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

  • zkfcfbzr@lemmy.worldEnglish
    4·
    7 months ago
    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)_/¯