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I need some help, any Ideas?

in L'Hospital's rule, replace the assumption that $\frac{f}{g}$ tends to $\frac{0}{0}$ with the assumption that it tends to $\frac{\infty}{\infty}$. if $\frac{f'}{g'}$ tends to $L$. prove that $\frac{f}{g}$ tends to $L$ also.

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    This might help, $$\kappa \xleftarrow{x\to\infty} \frac{f}{g}=\frac{1/g}{1/f}\;\xrightarrow{LH}\;\frac{-g\,'/g^2}{-f\,'/f^2}\xrightarrow{x\to\infty}\frac{\kappa^2}{L}.$$ I can't seem to show $\kappa\ne0$ when $L\ne0$.2012-02-03
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    @anon, could you please suggest a source I can refer to to know more about the formula you have given in the above comment? Thanks.2012-02-03
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    @Emmad: The LH stands for L'Hospitals, and I went from one side of it to the other by using the reciprocal differentiation rule. After that I just noted $g'/f'=(f'/g')^{-1}\to L^{-1}$ and flipped the $g^2$ and $f^2$ likewise to get $\kappa^2$. (The beginning of the formula defines $\kappa$ as $\lim(f/g)$)2012-02-03
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    @anon, pretty fancy!, thank you.2012-02-03
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    I don't think there is a simple reduction to the first case. You should rather try to show it without refering to the original result2012-02-03
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    Related: http://math.stackexchange.com/questions/100055/proof-of-lhopitals-rule-that-minimizes-special-casing2012-02-18
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    As @Listing wrote, there are *two* theorems called De l'Hospital' theorem. The case $[\infty/\infty]$ is not included in the case $[0/0]$.2012-07-04

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