L'Hopsital Rule Understanding Teacher taught us that if we ran into trouble finding the limit of a function we can take the power rule of derivative on the top and button of a fraction and then find the limit again.
But I don't understand, what does limit has to do with derivatives when the limit can't even be found?  Can someone explain it to me?
So if I can't find the limit of $\displaystyle \frac{x^3}{x^5}$ approaches infinity I can take the derivative of top and bottom until I can?  But why is that?  Isn't derivative just some kind of slope and limit is finding the missing point in graph?  what do they have to do with each other?
 A: I can give you a proof for the case when $x \to a$, but please don't take this proof seriously because it's a naive one.
So, suppose that $\displaystyle \lim_{x \to a} f(x)=0$ and $\displaystyle \lim_{x \to a} g(x)=0$. So if you want to find $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}$ you can do as follows:
$$\large\lim_{x \to a} \frac{\frac{f(x)-0}{x-a}}{\frac{g(x)-0}{x-a}}=\lim_{x \to a} \frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}=\frac{\lim_{x \to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x \to a}\frac{g(x)-g(a)}{x-a}}=\frac{f'(a)}{g'(a)} $$
In the first step, we add $0$ to the numerator and denominator, which is arithmetically fine because $0$ doesn't change anything when it's summed. Then we divide both the numerator and denominator by $x-a$ which is again OK because it's like you're multiplying by $1$ and multiplying by $1$ doesn't change anything. Then I'm using the theorem that $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}=\frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}$ which is again fine and at the end I use the definition of $f'(a)$
Now, notice that if $\displaystyle \lim_{x \to \infty}f(x)=\lim_{t \to 0}f(t)$ where $\displaystyle t= \frac{1}{x}$. So you can generalize what I said with a suitable substitution.
Anyway, my proof isn't really rigorous for an analyst, but I guess it gives you an idea of what's going on without using any further theorems like Mean Value Theorem.
A: Let's try to find the limit
$$
\lim_{x \to 0} \frac{\sin x}{x^3 + x}
$$
by direct substitution, we get $\frac{0}{0}$, which is an indeterminate form. So we use l'Hopital's rule and get $\frac{\cos x}{1 + 2x^2}$, which we can directly evaluate to be $1$.
Let's take a closer look at what just happened by writing using the taylor expansion. Our original limit then becomes
$$
\lim_{x \to 0} \frac{0 + x - \frac{x^3}{6} + \cdots}{0 + x + x^3}
$$
our first approximation, $\frac{0}{0}$, is sort of like using the zeroth-order approximation $\sin x \approx 0$ and $x^3 + x \approx 0$ for small $x$. Of course it's true, but it's not powerful enough. When we use the slightly more powerful first-order approximation $\sin x \approx x$ for small $x$ and $x^3 + x \approx 0$ for small $x$. Now this approximation is more accurate and powerful enough to evaluate the limit.
By differentiating, we replace the original limit with
$$
\lim_{x \to 0} \frac{1 - \frac{x^2}{2} + \cdots}{1 + 3x^2}
$$
notice how what used to be the first-order terms became zeroth order terms. Every iteration of l'Hopital's rule destroys the zeroth-order term, allowing the higher-order terms to finally be heard.
A: Roughly speaking this is because when you are dealing with limit of $f(x)/g(x)$ as $x \to x_0$ and $f(x_0) = g(x_0) = 0$, $x$ approaches $x_0$ very close, so $x-x_0$ is becoming very small and $f(x)-f(x_0)$ is also becoming small, so yes, you can go along the tangent line with the slope $f'(x)$:
$$f(x) = f(x0) + f'(x_0) (x-x_0) + ...$$
$$g(x) = g(x_0) + g'(x_0)(x-x_0) + ...$$
Don't forget you have $f(x_0) = g(x_0) = 0$.
If you are dealing with limit at $f(x), g(x) \to \infty$, you should rewrite
$$ \frac{f(x)}{g(x)} = \frac{1/g(x)}{1/f(x)} $$
This is of course not a strict proof
