# Understanding L'Hospital's Rule

In my lecture notes:

Why is $\lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f(x) - f(0)}{g(x)-g(0)}$ and so on. I know its trying to get to "$\frac{\text{change in y}}{\text{change in x}}$" but can I actually add stuff like that?

-

Since $f(0)=0$, it follows that $f(x) = f(x)-f(0)$. Similarly for $g(x)$.

Also, for any number $a$, and any number $b \ne 0$, we have that $a = a\cdot \frac{b}{b}$. Hence,

$\frac{f(x) - f(0)}{g(x)-g(0)} = \frac{f(x) - f(0)}{g(x)-g(0)} \frac{\frac{1}{x-0}}{\frac{1}{x-0}} = \frac{\frac{f(x)-f(0)}{x-0}}{\frac{g(x)-g(0)}{x-0}}$

-

This is not strictly an application of L'Hopital's rule, but rather (I think) a warm up to the proof of the rule for a specific case.

As anonymous points out above, given any expression, you may:

$\ \ \$1) add 0 to it, without changing it

$\ \ \$2) multiply it by 1 without changing it

In your problem, starting with the expression $$\tag{1} {1-\cos x\over x-x^2}$$ and setting $f(x)=1-\cos x$ and $g(x)=x-x^2$, we have (as noted in your notes) $f(0)=0=g(0)$. So, using 1), the expression $(1)$ can be written $${(1-\cos x) - (1-\cos 0)\over (x-x^2)- 0}$$ and then using 2) $$\tag{2} {(1-\cos x) - (1-\cos 0)\over (x-x^2)- 0}\cdot{x-0\over x-0} ={ {(1-\cos x) - (1-\cos 0)\over x-0 }\over{ (x-x^2)- 0 \over x-0}}$$ In terms of $f$ ang $g$, the right hand side of $(2)$ can be written as $$\tag {3} {f(x)-f(0)\over x-0}\over{g(x)-g(0)\over x-0}$$ (note the second to last line of the first bullet point in your notes has an typo)

So, since $(1)$ and $(3)$ are the same $$\lim_{x\rightarrow0}{1-\cos x\over x-x^2} =\lim_{x\rightarrow0}{{f(x)-f(0)\over x-0}\over{g(x)-g(0)\over x-0}} ={\lim\limits_{x\rightarrow0}{f(x)-f(0)\over x-0}\over\lim\limits_{x\rightarrow0}{g(x)-g(0)\over x-0}}$$

Now here's the important observation: the limits on the right hand side of the above are by definition derivatives. The whole point of doing the above manipulations was to arrive at this stage.

-