# Does anyone know l'Hôpital's rule for limits?

I was doing an assignment and the middle 2 questions are l'Hôpital's rule questions and we haven't even done this in class yet. These are 1 or 2 chapters away yet we are expected to do them and I have no idea how to do these because we haven't learned yet so I was hoping someone here can help me.

First question:

An incorrect use of l'Hôpital's rule is illustrated in the following limit computation. Explain what is wrong and find the correct value of the limit.

$$\lim_{x\to\pi/2} \frac{\sin(x)}{x} = \lim_{x\to\pi/2} \frac{\cos(x)}{1}=0$$

Second question is:

Find $$\lim\limits_{x\to-\infty} xe^x$$ using l'Hôpital's rule. Using this information does the function $$xe^x$$ have a horizontal or vertical asymptote? state the equation of the asymptote.

I have done about 15 pages worth of questions and some I knew because I studied extra things on spare time but these are just hard.

• Are you sure you have that second one right? I don't think you can use L'Hopital's rule on that one. L'Hopital's rule can only be used if the numerator and denominator each approach 0 or if they each approach plus or minus infinity, referred to as an indeterminant form. If that second one were $xe^{-x}$, it could be rewritten as $\frac x{e^x}$, in which case both top and bottom approach infinity. L'Hopital's rule would then allow you to differentiate the top and bottom. – Mike Dec 1 '11 at 13:49
• Questions about what you have not done in class ... you should see the instructor about that, not us! – GEdgar Dec 1 '11 at 14:36
• @Mike: There was a minus sign that got lost in an edit: The second problem was $\lim\limits_{x \to -\infty} x e^{x}$. – t.b. Dec 1 '11 at 14:36

I would advise looking ahead in your book and just check out what l'Hôpital's rule says. This site is useful for answering a specific question, and helpful to use as a supplement to learning in class or from a book, but not really designed with teaching of a broader subject. I would also check with your instructor, as it is possible that those two questions were mistakenly included in this assignment.

That said, from Wikipedia's entry:

In its simplest form, l'Hôpital's rule states that for functions $f$ and $g$: $$\text{If }\lim_{x\to c} f(x)=\lim_{x\to c} g(x)=0 \text{ or } \pm \infty\text{ and }\lim_{x\to c} \frac {f'(x)}{g'(x)} \text{ exists,}$$ $$\text{then } \lim_{x\to c} \frac {f(x)}{g(x)}=\lim_{x\to c} \frac{f'(x)}{g'(x)}\ .$$

From this, you should be able to see why the application of l'Hôpital's is incorrect to apply to $\lim_{x\to \pi/2}\sin(x)/x$. What is the requirement of the rule? Does your limit satisfy that requirement?

For the second question, you can proceed as follows:

$$\lim_{x\to -\infty} x e^x = \lim_{x\to \infty} -x e^{-x} = \lim_{x\to\infty} \frac {-x}{e^x}\ ,$$

which then succumbs to l'Hôpital's rule.

Again, I would encourage you to check with your professor on these questions. If you haven't covered l'Hôpital's rule yet, my guess is that they did not mean to include them in this assignment.

• There was a minus sign that got lost in an edit. The second problem was $\lim\limits_{x \to -\infty} x e^{x}$. – t.b. Dec 1 '11 at 14:36
• @t.b. Thanks, I updated my answer. – process91 Dec 1 '11 at 15:11

l'Hospital's rule only applies for $$\lim \frac{f(x)}{g(x)}$$ if $\lim f(x)$ and $\lim g(x)$ is $\pm\infty$, or if $\lim f(x) = \lim g(x) = 0$.

And if they don't, then the limit is trivially found to start with, as you can see in $$\lim_{x\to\pi/2} \frac{\sin(x)}{x} = \frac{\sin(\pi/2)}{(\pi/2)} = \frac{2}{\pi}$$ Neither nominator or denominator goes to zero or infinity.

For the second question, are you sure you got it right? l'Hospital doesn't apply as it reads right now. Edit: The sign was lost in the edit when i answered this. For the current expression, l'Hospital holds, and its straightforward to use it.

• I guess you are supposed to rewrite the second expression to apply it... – Listing Dec 1 '11 at 13:50
• @Listing Even after rewriting it, the question doesn't make sense, since it asks to "state the equation of the asymptote". – process91 Dec 1 '11 at 13:53
• Yea I see, it is probably $xe^{-x}$ :) – Listing Dec 1 '11 at 14:10
• There was a minus sign that got lost in an edit. The second problem was $\lim\limits_{x \to -\infty} x e^{x}$ – t.b. Dec 1 '11 at 14:36
• It is incorrect to say the rule "only applies" in said cases, since in the first case one doesn't need that the numerator $\to \pm\infty\:.\:$ See this post for a precise statement (from Rudin's PoMA). – Bill Dubuque Dec 1 '11 at 18:52

$$\lim_{x\to-\infty} xe^x = \lim_{x\to-\infty} \frac{x}{e^{-x}}.$$ The numerator approaches $-\infty$ and the denominator approaches $\infty$, so L'Hopital's rule is applicable.

(The other parts of your question seem to have been dealt with already.)

• But one doesn't need the the numerator $\to \pm\infty$ to apply the rule in this case. See this post for a precise statement (from Rudin's PoMA). – Bill Dubuque Dec 1 '11 at 18:55
• @BillDubuque Very cool, I did not know that! Although, presumably, the only cases where l'Hopital's rule would be needed are those in which the numerator $\to \pm \infty$, since, if the numerator is bounded, then the limit would be 0. – process91 Dec 1 '11 at 18:57
• @process91 E.g. consider a case where the numerator is so complex that one cannot easily deduce its limit. – Bill Dubuque Dec 1 '11 at 19:42
• @BillDubuque True, that could be useful. Thanks! – process91 Dec 1 '11 at 20:34