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I was trying to prove L'Hôpital's rule for $a \to \infty$ and I came up with the following formula $$\lim _{x\to \:a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\lim _{x\to a}\left(\frac{g'\left(x\right)f\left(x\right)^2}{g\left(x\right)^2f'\left(a\right)}\right)$$ Can I prove L'Hôpital's rule from here? I have tested it for a couple of limits and it seems to work.

These were my steps, let $b=\frac{1}{f\left(x\right)}$ and $c=\frac{1}{g\left(x\right)}$

$ \to \lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\lim _{x\to a}\left(\frac{\frac{1}{g\left(x\right)}}{\frac{1}{f\left(x\right)}}\right)=\lim _{x\to a}\left(\frac{c\left(x\right)}{b\left(x\right)}\right)$

Assuming $a$ is infinity, we can use a linear approximation method like this:

$b(x)≈b′(a)(x−a)+b(a)$

$c(x)≈c′(a)(x−a)+c(a)$.

So we get the above result when we differentiate $b$ and $c$ and substitute there original values of $f(x)$ and $g(x)$. Where did I go wrong?

  • If $a=\infty$, to you really expect an approximation featuring an $x-a$ factor to work? – J.G. Oct 20 '23 at 07:17
  • I guess that doesn't make sense. I was just following the comments under J126's answer for this post: https://math.stackexchange.com/questions/584889/the-intuition-behind-lhopitals-rule. Surprisingly enough the top result seems to be true. –  Oct 20 '23 at 07:26
  • Thanks for that context. See @JohnAlexiou's comment underneath that answer. – J.G. Oct 20 '23 at 07:30
  • @J.G. I am not really sure what he means by transformation. Can you elaborate a bit on what that means? I was following chubakueno's comment right underneath that. –  Oct 20 '23 at 07:32
  • @PaulFrost Yes it does. How did you find a link with the exact question? –  Oct 20 '23 at 07:34
  • @OmarAboutaleb Because I remembered that I have answered it ;-) – Paul Frost Oct 20 '23 at 10:21

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