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Suppose $f$ and $g$ are differentiable and $g'(x)\neq 0$ on an open interval $I$ that contains $a$.

Suppose that $$\lim_{x\to a}f(x)=L=\lim_{x\to a}g(x)\quad L=0\,or\,\infty\,or-\infty$$, then $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$

I proved the case when $L=0$,and I wrote a proof about the case when $L=\infty$ ,I found it is different from the proof on the internet, so I don't know whether it is right.

My proof: $$\lim_{x\to a}f(x)=\infty=\lim_{x\to a}g(x)$$ $$\lim_{x\to a}\frac{1}{f(x)}=0=\lim_{x\to a}\frac{1}{g(x)}$$ $$\lim_{x\to a}\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}=\lim_{x\to a}\frac{[\frac{1}{g(x)}]'}{[\frac{1}{f(x)}]'}$$.

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{g'(x)[f(x)]^2}{f'(x)[(g(x)]^2}$$ $$\lim_{x\to a}\frac{f(x)}{g(x)}[1-\frac{f(x)g'(x)}{g(x)f'(x)}]=0$$.

I think L'Hospital assumes that $\lim_{x\to a}\frac{f(x)}{g(x)}$ exist, or is $\infty$, or $-\infty$, and so $$\lim_{x\to a}\frac{f(x)g'(x)}{g(x)f'(x)}=1$$ $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$.

If it is wrong, please tell me why. Thank you very much!

Yang
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    Maybe in the least second and least third lines it's needed to show that limit, that you carry out of the $u(x)\cdot v(x)$ multiplication, exists. If somehow proven that the limits exist, I like the proof. – Alexey Burdin May 25 '20 at 02:41
  • @AlexeyBurdin Thank you ,I think I am wrong .because if $\lim \frac{f(x)}{g(x)}=0$,I can't deduce the follow – Yang May 25 '20 at 02:53
  • There is a fundamental misconception here. The existence of limit of $f'/g'$ is a part of the hypotheses and then the conclusion is that $f/g$ also tends to same value. Further the second form is "$\text{anything} /\pm\infty$" and we don't worry about numerator here. – Paramanand Singh May 27 '20 at 07:42
  • Now rewrite the proof based on the assumption that $f'/g'$ tends to a limit. – Paramanand Singh May 27 '20 at 07:43

1 Answers1

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This step here

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{g'(x)[f(x)]^2}{f'(x)[(g(x)]^2}$$ $$\lim_{x\to a}\frac{f(x)}{g(x)}[1-\frac{f(x)g'(x)}{g(x)f'(x)}]=0$$

doesn't work because the limit in the first line could be $\infty$. If $\lim_{x\rightarrow a} p(x) = \infty$ and $\lim_{x\rightarrow a} q(x) = \infty$, then you can make no conclusions about $\lim_{x\rightarrow\infty} p(x) - q(x)$. Similarly, this step $$\lim_{x\to a}\frac{f(x)g'(x)}{g(x)f'(x)}=1$$ $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ doesn't work because $\lim_{x\rightarrow a} f(x)/g(x)$ could be $0$.

You might be able to patch up the proof by considering the $0$ and $\infty$ limits as special cases, as it does seem to work if it is known that the limit is finite and nonzero.

eyeballfrog
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