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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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    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

1 Answers 1

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Suppose $f$ and $g$ are continuous and differentiable for all $x\neq a$, and that g' is never zero. Suppose that $\mathop {\lim }\limits_{x \to a } f\left( x \right) = \mathop {\lim }\limits_{x \to a } g\left( x \right) = \infty $

and that the limit

\mathop {\lim }\limits_{x \to a } \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = L

exists. Then \mathop {\lim }\limits_{x \to a } \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = \mathop {\lim }\limits_{x \to a } \frac{{f\left( x \right)}}{{g\left( x \right)}} = L

According to Cauchy we have ($\alpha < x < a$ and $\alpha < c < x$ )

\frac{{f\left( x \right) - f\left( \alpha \right)}}{{g\left( x \right) - g\left( \alpha \right)}} = \frac{{f'\left( c \right)}}{{g'\left( c \right)}}

Let's write the first member of the equality as

\frac{{f\left( x \right)}}{{g\left( x \right)}}\frac{{1 - \frac{{f\left( \alpha \right)}}{{f\left( x \right)}}}}{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}} = \frac{{f'\left( c \right)}}{{g'\left( c \right)}}

Then

\frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{f'\left( c \right)}}{{g'\left( c \right)}}\frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( \alpha \right)}}{{f\left( x \right)}}}}(\star)

From the condition \mathop {\lim }\limits_{x \to \infty } \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = L we get that for any $\epsilon > 0$, we can choose $\alpha$ arbitrarely close to $a$ such that for all $x=c$ where $\alpha < c < a$, we have

\left| \frac{{f'\left( c \right)}}{{g'\left( c \right)}}-L\right| < \epsilon

L-\epsilon <\frac{{f'\left( c \right)}}{{g'\left( c \right)}} < L+\epsilon

Now, since

$\mathop {\lim }\limits_{x \to a} \frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( a \right)}}{{f\left( \alpha \right)}}}} = 1$

We have

$1-\epsilon < \frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( a \right)}}{{f\left( \alpha \right)}}}} < 1+\epsilon$

Multiplying the previous inequalities gives

\left( {L - \epsilon} \right)\left( {1 - \epsilon} \right) < \frac{{f'\left( c \right)}}{{g'\left( c \right)}}\frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( a \right)}}{{f\left( \alpha \right)}}}} < \left( {L + \epsilon} \right)\left( {1 + \epsilon} \right)

and because of $(\star)$ we get

$\left( {L - \epsilon} \right)\left( {1 - \epsilon} \right) < \frac{{f\left( x \right)}}{{g\left( x \right)}} < \left( {L + \epsilon} \right)\left( {1 + \epsilon} \right)$

Since $\epsilon$ is arbitrarely small, for $x$ sufficiently close to $a$ we will have

$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = L$