TITLE: Proof of L'Hospitals Rule
QUESTION [17 upvotes]: No matter where I would look it would seem that L'Hospital's Rule has a strange proof-given that they teach it in high school, it seems troublesome that I can't find a solid proof at that level of knowledge. Does anyone have a proof that is fairly basic? Or does proving it simply require higher math?

REPLY [20 votes]: One can prove it using linear approximation:
Recall the definition of the derivative, $f'(x)$:
$$
  f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}
$$
which we can write as
$$
    f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h)
$$
where $\eta(h)$ is a function such that $\lim_{h\rightarrow 0} \eta(h)=0$ (and is continuous).
Then in the above displayed equation multiply through by $h$, which gives (after rearrangment and switching $\eta(h)$ with $-\eta(h)$)
$$
f(x+h) = f(x) +f'(x)h+h\cdot \eta(h).
$$
Now use L'Hopitals rule for the limit of $f(x)/g(x)$ when $x$ goes to $x_0$, so we have 
$f(x_0)=g(x_0)=0$. Then by using linear approximation as above in both numerator and denominator (and writing the $\eta$-function for $g$ as $\epsilon$), we get
$$
\frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)}
$$
and now, in both numerator and denominator, there is a common factor $h$ we can cancel. Taking the limit as $h \rightarrow 0$ now gives the result. 

This proof is an upgraded version of Bernoulli's original one. See https://mathoverflow.net/a/51691