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 $-\frac1{2}\neq \frac 1{24}$.

Solution 1:

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".

Solution 2:

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., $\dfrac 00, \;\text{or}\;\dfrac {\pm\infty}{\pm\infty}$.

Solution 3:

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; $$\lim_{x\to\infty} \frac{x}{x+\sin x} = \lim_{x\to\infty} \frac{1}{1+\frac{\sin x}{x}}=1.$$

Solution 4:

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.