# What limits L'Hôpital's rule?

I have very often heard that L'Hospital's rule can only be applied when we have $$0 \over 0$$ or $$\infty \over \infty$$ form in limit.

But what I do no know is what prevents us from applying it in case of other forms of limits like of type $$1^\infty$$.

I reckon it has something to do with how L'Hospital's rule is derived but I am not sure.

Question : what actually prevents the usage of L'Hospital's rule to all cases of limits?

• Why on earth would it work? What you should ask yourself first is why it does work in some cases. Then it should hopefully become obvious why it doesn't work in other cases. – Hans Lundmark Mar 28 '20 at 12:03
• That is what exactly what I wanted to know and so I asked this question. – Hrishabh Nayal Mar 28 '20 at 12:23
• You may perhaps know that there are theorems which involve hypotheses and thus their conclusion holds only under those hypotheses. A proper statement of the L'Hospital's Rule mentions the hypotheses under which it holds. One should not use any result without knowing its actual statement / formulation. – Paramanand Singh Mar 29 '20 at 2:46

We can't apply L'Hospital's Rule directly in that situation because it doesn't work.

For instance, If $$f(x)=1+\frac1x$$ and $$g(x)=x$$, then $$\lim_{x\to\infty}f(x)^{g(x)}=e$$. However,$$\lim_{x\to\infty}f'(x)^{g'(x)}=\lim_{x\to\infty}-\frac1{x^2}=0\neq e.$$

• Yes, but why does it not work in $some$ cases? – Hrishabh Nayal Mar 28 '20 at 11:32
• Did you ever read the proof of L'Hopital's Rule? It is essential for it that you are dealing with quotients. – José Carlos Santos Mar 28 '20 at 11:34
• Sorry, Not to sound like a broken record but why quotients? If you could give me link where I could read the proof(or give it yourself) it should help. Thanks – Hrishabh Nayal Mar 28 '20 at 11:40
• There are several ones here, for instance. – José Carlos Santos Mar 28 '20 at 11:51
• Oh thanks! I get now why we need quotients thanks. – Hrishabh Nayal Mar 28 '20 at 12:27

For example of limit of the form $$\;1^\infty\;$$, take the classic

$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=\lim_{x\to\infty}e^{x\log\left(1+\frac1x\right)}$$

So, by continuity of the exponential function, you need to evaluate

$$\lim_{x\to\infty} x\log\left(1+\frac1x\right)=\lim_{x\to\infty}\frac{\log\left(1+\frac1x\right)}{\frac1x}$$

and the last one is just a $$\;\frac00\;$$ form limit which you can solve easily with L'Hospital...!

The above example is classic and something similar can be applied to most $$\;1^\infty\;$$ limits used in Calculus I courses

• What I wanna know is why you cant apply it directly to another form and need to convert it. – Hrishabh Nayal Mar 28 '20 at 11:33
• Because the other form is neither $\;\frac\infty\infty\;$ nor $\;\frac00\;$ . Simple. – DonAntonio Mar 28 '20 at 11:39
• Sorry I think I should have asked better. What I want to know is why L'Hôpital's rule fails in cases like 1/$\infty$ – Hrishabh Nayal Mar 28 '20 at 11:42
• Do you know the proofs (at least the simplest one) of the L-H theorem? Then you can see that the case $\;\frac1\infty\;$ doesn't work. It's like asking "why only a straight triangle fulfills Pythagoras theorem? Well, because ! – DonAntonio Mar 28 '20 at 12:09
• I get it now thanks! I have seem some proofs and it has completely clear thanks for your answer – Hrishabh Nayal Mar 28 '20 at 12:30