# Using L'Hospital rule for any quotient

Why can't L'Hospital rule be used to compute the limit of any quotient rather that just indeterminate forms.

For example: why can't we evaluate the limit of $\lim\limits_{y \rightarrow 0} \frac{y^2 + 1}{y} = \lim\limits_{y \rightarrow 0} 2y = 0$.

Why does it only apply to indeterminate forms?

• @Regret It does not work in the example ($\lim\limits_{y \rightarrow 0}\frac{y^{2}+1}{y}$) I gave above either, since the limit does not exist for the example I gave. I am interested in why the rule only applies to indeterminate forms.
– user116403
Commented Mar 9, 2015 at 9:55
• I totally missed that you had an example where it fails. Sorry. I am quite off today. Commented Mar 9, 2015 at 9:56
• @Regret No prob.
– user116403
Commented Mar 9, 2015 at 9:58

Recall that the derivative of a function at a point $a$ is the best linear approximation. Roughly, this means that very near some constant $a$, we can write
$$f(x) = f(a) + f'(a)(x-a) + \text{very small stuff}$$ where the small stuff is negligible if $x$ is close to $a$. Now suppose you have $g(x)$ (for which you can also write a linear approximation) and you want to evaluate $$\lim_{x \to a} \frac{f(x)}{g(x)}.$$ When we take the limit, we only care about values where $x$ is close to $a$, so we can ignore the small stuff (if you want more precision, ask in the comments, but I hope you get the idea). Therefore $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{ f(a) + f'(a)(x-a)}{ g(a) + g'(a)(x-a)}.$$ Now, if $f$ and $g$ both vanish at $a$, i.e., if you started with an "indeterminate form" $0/0$, then the right hand side is $$\lim_{x \to a} \frac{ 0 + f'(a)(x-a)}{ 0 + g'(a)(x-a)} = \frac{f'(a)}{g'(a)}$$ and you recover L'Hôpital's rule ! On the other hand, if $f$ and $g$ don't both vanish at $a$, we don't get the simplification, and so you have to find some other way to evaluate the limit (but usually it's easy, since if $g$ vanishes at $a$ and $f$ doesn't the limit won't exist, and if $f$ vanishes at $a$ and $g$ doesn't, the limit will be zero. If neither vanishes then the limit is just the quotient of the values.)
• (It should perhaps be mentioned that this proof only covers the case when $g'(a)\neq 0$; the general case is trickier.) Commented Mar 9, 2015 at 11:31