L'Hospital's Rule of infinity over infinity

Question

I am troubled for understanding the L'Hospital's Rule of $\infty/\infty$ :

$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}= \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)} \tag{1}$$

where $\lim_{x\rightarrow a}\{f(x),g(x)\} \rightarrow \infty$.

The equation (1) is well understood for me in the case of $\lim_{x\rightarrow a}\{f(x),g(x)\} \rightarrow 0$, in which case :

$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{\Delta x\rightarrow 0}\frac{f(a)+f^\prime(a)\Delta x}{g(a)+g^\prime(a)\Delta x}= \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)} \tag{2}$$

However, if $\lim_{x\rightarrow a}\{f(x),g(x)\} \rightarrow \infty$, here is my attempt :

$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{1/g(x)}{1/f(x)}= \lim_{x\rightarrow a}\frac{-\frac{1}{g^2(x)}g^\prime(x) }{-\frac{1}{f^2(x)}f^\prime(x)} \tag{3}$$

I don't understand why this last result can be turned into $\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}$?

Answer 1

Here is the Intuition behind that.
It lets you see the Conclusion in easy terms.

View 1 :

You have got this :
$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{1/g(x)}{1/f(x)}= \lim_{x\rightarrow a}\frac{-\frac{1}{g^2(x)}g^\prime(x) }{-\frac{1}{f^2(x)}f^\prime(x)} \tag{3}$

Now , let $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=L \tag{4}$

Then , we have $L= \lim_{x\rightarrow a}\frac{-\frac{1}{g^2(x)}g^\prime(x) }{-\frac{1}{f^2(x)}f^\prime(x)}=\lim_{x\rightarrow a}\frac{f^2(x)}{g^2(x)}\frac{g^\prime(x)}{f^\prime(x)}=\lim_{x\rightarrow a}L^2\frac{g^\prime(x)}{f^\prime(x)} \tag{5}$

Hence $\frac{1}{L}=\lim_{x\rightarrow a}\frac{g^\prime(x)}{f^\prime(x)} \tag{6}$

Hence $L=\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)} \tag{7}$

Alternate View 2 :

We have $L= \lim_{x\rightarrow a}\frac{-\frac{1}{g^2(x)}g^\prime(x) }{-\frac{1}{f^2(x)}f^\prime(x)}=\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\frac{f(x)}{g(x)}\frac{g^\prime(x)}{f^\prime(x)}=\lim_{x\rightarrow a}L\frac{f(x)}{g(x)}\frac{g^\prime(x)}{f^\prime(x)} \tag{8}$

The last term must be $1$ , hence that must involve $L$ times the reciprocal.

$1=\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\frac{g^\prime(x)}{f^\prime(x)} \tag{9}$

OBSERVATION :

No other value for the given term will be Consistent , over-all.
We can show that this value is Correct & valid though other ways , though this is the Intuitive way.

Answer 2

If you are comfortable with $\frac{0}{0}$, then I would do the following:

1. $\frac{B}{A} = \frac{1}{\frac{A}{B}}$
2. In limits, you usually do $0 = \frac{1}{\infty}$
3. Let us say that our limit $L = \frac{f(x)}{g(x)} = \frac{\infty}{\infty}$. You can also express $\frac{\infty}{\infty} = \frac{1}{\infty}\cdot\frac{1}{\frac{1}{\infty}} = \frac{\frac{1}{\infty}}{\frac{1}{\infty}} = \frac{0}{0}$.
4. Because this is also a form of $\frac{0}{0}$ then L'Hospital's rule applies.

Answer 3

You idea does not help you to prove L'Hospital's Rule in the $\infty/\infty$-case. You only get the formula $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{1/g(x)}{1/f(x)}= \lim_{x\rightarrow a}\frac{-\frac{1}{g^2(x)}g^\prime(x) }{-\frac{1}{f^2(x)}f^\prime(x)} = \lim_{x\rightarrow a}\frac{f^2(x)}{g^2(x)} \cdot \frac{g^\prime(x) }{f^\prime(x)} $$ provided the limit on the RHS exists. But this does not follow from the existence of $\lim_{x\rightarrow a}\dfrac{f'(x)}{g'(x)}$. We have to know additionally the existence of $\lim_{x\rightarrow a}\frac{f^2(x)}{g^2(x)}$ which is up to sign the same as the existence of $\lim_{x\rightarrow a}\dfrac{f(x)}{g(x)}$.

Only in that case we can deduce that $\lim_{x\rightarrow a}\dfrac{f(x)}{g(x)}=\lim_{x\rightarrow a}\dfrac{f'(x)}{g'(x)}$ which is at best a computational help if we know that both limit exist. But the point of L'Hospital's Rule is that it suffices to know that $\lim_{x\rightarrow a}\dfrac{f'(x)}{g'(x)}$ exists.

Indeed you need a separate proof in the $\infty/\infty$-case. It is not reducible to the $0/0$-case.