2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ @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. $\endgroup$
    – user116403
    Commented Mar 9, 2015 at 9:55
  • $\begingroup$ I totally missed that you had an example where it fails. Sorry. I am quite off today. $\endgroup$
    – Regret
    Commented Mar 9, 2015 at 9:56
  • $\begingroup$ @Regret No prob. $\endgroup$
    – user116403
    Commented Mar 9, 2015 at 9:58

2 Answers 2

Reset to default
5
$\begingroup$

Sounds like you know that L'Hôpital's rule won't work and are asking why. So let's see why L'Hôpital's rule is true.

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.)

$\endgroup$
2
  • 1
    $\begingroup$ (It should perhaps be mentioned that this proof only covers the case when $g'(a)\neq 0$; the general case is trickier.) $\endgroup$
    – Hans Lundmark
    Commented Mar 9, 2015 at 11:31
  • $\begingroup$ @HansLundmark good point! I'll edit it in if OP cares. $\endgroup$
    – hunter
    Commented Mar 9, 2015 at 11:41
2
$\begingroup$

I think you are asking the wrong type of question. It's a little like when you ask why can't I use linearity when evaluating a square root? The real question to ask is how come l'Hospitals rule does work for indeterminate forms?

Even so an easy answer to your question is because in general the quotient of derivatives of two functions has little to do with the quotient of their values. This can easily be seen when you realize that adding a constant to a function does not change it's derivative, but obviously changes it's value (and thus it's limit).

$\endgroup$

You must log in to answer this question.