When to Stop Using L'Hôpital's Rule I don't understand something about L'Hôpital's rule. In this case: 
$$
\begin{align}
& {{}\phantom{=}}\lim_{x\to0}\frac{e^x-1-x^2}{x^4+x^3+x^2} \\[8pt]
& =\lim_{x\to0}\frac{(e^x-1-x^2)'}{(x^4+x^3+x^2)'} \\[8pt]
& =\lim_{x\to0}\frac{(e^x-2x)'}{(4x^3+3x^2+2x)'} \\[8pt]
& =\lim_{x\to0}\frac{(e^x-2)'}{(12x^2+6x+2)'} \\[8pt]
& = \lim_{x\to0}\frac{(e^x)'}{(24x+6)'} \\[8pt]
& = \lim_{x\to0}\frac{e^x}{24} \\[8pt]
& = \frac{e^0}{24} \\[8pt]
& = \frac{1}{24}
\end{align}
$$
Why do we have to keep on solving after this step:
$$\displaystyle\lim_{x\to0}\dfrac{(e^x-2)'}{(12x^2+6x+2)'}$$
Can't I just plug in $x=0$ and compute the limit at this step giving me:
$$\dfrac{1-2}{0+0+2}=-\dfrac{1}{2}$$
I'm very confused, because I get different probable answers for the limit, depending on when do I stop to differentiate, as clearly $-\frac1{2}\neq \frac 1{24}$.
 A: One must be very careful about using l'Hospital's rule.  It applies only when the numerator and denominator both tend to $0$ or $\infty$ and the denominator is never $0$ in a punctured neighborhood of the point.  A denominator of the form $x\sin \frac1x$ is ineligable for the rule.
But, students should try to avoid the rule anyway.  Here is a parable.  A student is assigned the task of finding
$$
\lim_{x\to 0} \frac{\sin^6 x}{x^6}.
$$
A bad student cancels the $6$ and the $x$ giving $\sin$.
A naive student applies l'Hospital's rule 6 times and eventually gets $\frac{720}{720} = 1$.
A mediocre student applies the rule once, and gets 
$$
\lim_{x\to 0} \frac{6\sin^5 x \cos x}{6x^5}.
$$
He cancels the $6$, and removes the $\cos x$ term since it tends to $1$.  He repeats the process 5 more times.
A good student writes the expression as:
$$
\lim_{x\to 0} \left[\frac{\sin x}{x}\right]^6.
$$
He uses continuity of $t^6$ to move the limit inside the brackets and gets $1^6 = 1$.
An engineer says "$\sin x = x$, so the expression is $1$."  And, he'd be right!

Using l'Hospital's rule can also hurt the students understanding of a problem.  Here's a classic example.
The $p$-power mean of two positive numbers, $x$ and $y$ is defined as:
$$
M_p(x,y) = \left[\frac{x^p + y^p}{p}\right]^{\frac1p}.
$$
Can one provide suggestive evidence without using l'Hospital's rule that
$$
\lim_{p\to0^{+}} M_p(x,y) = \sqrt{xy}?
$$
One way to do this is to use the approximation $x^p \approx 1 + p \log x$ for $p$ small, taken from the Taylor expansion of $x^p$ at $p=0$.  Substitute into the power mean formula, and one gets:
$$
\left[1 + \frac p2 (\log x + \log y)\right]^{\frac1p} = \left[1 +  p \log \sqrt{xy}\right]^{\frac1p}.
$$
Let $s = \frac1p$, and this is 
$$
\left[1 +  \frac{\log \sqrt{xy}}{s}\right]^s
$$
The limit of that as $s\to{+\infty}$ is
$$
e^{\log\sqrt{xy}} = \sqrt{xy}.
$$
Making this rigorous is not trivial.
A: You can't use L'hopital in the third equality sign because you don't have a $"0/0"$ expression.
You get
$$
\lim_{x\to 0}\frac{e^x - 2x}{4x^3 + 3x^2 + 2x}.
$$
Here the numerator approaches $1$ and the denominator approaches $0$, so in all the limit doesn't exist.
Also note that
$$
4x^3 + 3x^2 + 2x 
$$
is negative when $x <0$ and it is positive when $x>0$. Therefore one can't say that the limit is $\infty$ or $-\infty$. The limit just doesn't exist.
A: Once your answer is no longer in the form 0/0 or $\frac{\infty}{\infty}$ you must stop applying the rule. You only apply the rule to attempt to get rid of the indeterminate forms. If you apply L'Hopital's rule when it is not applicable (i.e., when your function no longer yields an indeterminate value of 0/0 or $\frac{\infty}{\infty}$) you will most likely get the wrong answer. 
You should have stopped differentiating the top and bottom once you got to this:
$\dfrac{e^x-2x}{4x^2+3x^2+2x}$. Taking the limit at that gives you $1/0$. The limit is nonexistent. 
Also, don't be tempted to say "infinity" when you see a 0 in the denominator and a non-zero number in the top. It may not be the case. For example, the function $\frac{1}{x}$ approaches infinity and negative infinity from both sides of the limit as x approaches 0. Its not necessarily infinite; its best just to leave it as "nonexistent". 
A: After differentiating just once, you get $$\lim_{x \to 0} \dfrac{e^x-2x}{4x^3+3x^2+2x}$$ which "evaluates" to $\dfrac 10$, i.e., the numerator approaches $1$, and the denominator approaches $0$. Hence, L'Hopital no longer applies and we have $$\lim_{x \to 0} \dfrac{e^x-2x}{4x^3+3x^2+2x}\quad\text{does not exist}.$$  
L'Hopital's rule applies provided and only while a limit evaluates to an "indeterminate" form: e.g., $\dfrac 00, \;\text{or}\;\dfrac {\pm\infty}{\pm\infty}$.
A: From the second equality we find
$$\lim_{x\to0^+}\frac{e^x-2x}{4x^3+3x^2+2x}=\infty$$
thus the other equalities are false.
A: A quick addition to Ra1nMaster's otherwise excellent answer: you can only apply L'Hopital's rule if you have an indeterminate form and if the limit, after applying L'Hopital's rule, exists.
This second condition is equally important; for instance a classic stumper is
$$\lim_{x\to\infty} \frac{x}{x+\sin x}.$$
Since this limit has the form $\frac{\infty}{\infty}$, one might naively apply L'hopital's rule, getting
$$\lim_{x\to\infty} \frac{1}{1+\cos x}$$
and concluding the original limit does not exist. This is wrong;
$$\lim_{x\to\infty} \frac{x}{x+\sin x} = \lim_{x\to\infty} \frac{1}{1+\frac{\sin x}{x}}=1.$$
A: L'Hôpital's rule is ment to applied only when you have to limit of the form 0/0. Once your limit is not in 0/0 form, you are not supposed to apply L'Hôpital's rule.
A: In layman's terms:
L'Hôpital's rule should only be applied when we have the limit of the form:
0/0
Read: http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx
and
http://tutorial.math.lamar.edu/Classes/CalcI/TypesOfInfinity.aspx
Another tip:
If you are using the rule and you find yourself differentiating many times then there is another way to solve these limits. Like simple reduction or change of base.
