# another form of the L'Hospital's rule

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.

-
This might help, $$\kappa \xleftarrow{x\to\infty} \frac{f}{g}=\frac{1/g}{1/f}\;\xrightarrow{LH}\;\frac{-g\,'/g^2}{-f\,'/f^2}\xrigh‌​tarrow{x\to\infty}\frac{\kappa^2}{L}.$$ I can't seem to show $\kappa\ne0$ when $L\ne0$. – anon Feb 3 '12 at 7:59
@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. – NoChance Feb 3 '12 at 8:35
@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 $\lim(f/g)$) – anon Feb 3 '12 at 8:37
@anon, pretty fancy!, thank you. – NoChance Feb 3 '12 at 9:16
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 result – Listing Feb 3 '12 at 10:49

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

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

-