# When to use L'Hospital's rule vs the limit shortcut

Before being so quick to downvote or throw darts, please forgive my ignorance and inability to recall basic calculus atm.

Consider the limit $\lim \limits_{x \to \infty}\frac{3x^2+14x-5}{x^2+x-12}$. It can quickly be determined that the $x$ approaches $3$ using L'Hospital's rule.

The same answer can be derived, however, by using this shortcut (I am unaware if it has a specific name):

$\lim \limits_{x \to \infty}\frac{3x^2+14x-5}{x^2+x-12}=\lim \limits_{x \to \infty}\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}$

Every term is divided by the highest degree of $x$ in the denominator. The terms with $x^2$ in both the numerator and denominator simplify. All other terms go to $0$. Therefore:

$\lim \limits_{x \to \infty}\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}=\frac{3}{1}=3$

1) What are the explicit conditions in which may I use this shortcut?

2) How do I know when to use it over L'Hospital's rule, as both techniques can be used only when working with quotients?

3) What is the shortcuts name, if any?

• Your second method is the more common one --- you use L'Hopital's when it fails (provided you satisfy the right conditions for L'Hopital's). This typically works when you have a quotient of polynomials (or in cases like totoro has pointed out), whereas L'Hopital's works for other kinds of functions, too. I don't think it has a universal name. – Bill Wallis May 11 '18 at 21:35
• In class I usually recommend it when there is a quotient of sums of terms that can easily be compared in its growth to infinity. It would also be applicable, for example, to $\lim_{n\to\infty}\frac{n!+3^n-n^{50}}{2\ln(n)-e^{n^2}+2\sin(n)}$. Here 'compared' means knowing the limit of the quotient of the different terms. whether the quotient tends to $0$, to $\infty$, or if it remains bounded. – user551819 May 11 '18 at 21:37
• In my opinion L'Hopital is rarely the method of choice. Almost any other that works is more informative. This trick is good for quotients of polynomials. In general, writing down the first few terms of the power series for the various elements and looking at the leading term will often help. There are questions/answers on this site on the general theme "why not to use L'Hopital". – Ethan Bolker May 11 '18 at 21:40
• L'Hopital can also be used in not quotient indeterminate limits. – Namaste May 11 '18 at 21:42
• One can also use simplification of rational functions, e.g. $\lim_{x\to 2} \frac{x-2}{x^2-4} = \lim_{x\to 2} \frac 1{x+2} = \frac 14.$ Or $$\lim_{x\to 1}\frac{x^2+x-2}{x^3-x^2-3x+3} = \lim_{x\to 1}\frac{x+2}{x^2-3} = -\frac 32$$ – Namaste May 11 '18 at 21:51

Shorcut can be used in quotient limits anytime there exist some leading term at the numerator and denominator which becomes dominant in the limit with respect to the others terms that is, indicating with g(x) the dominant term

$$\lim \limits_{x \to \infty}\frac{f(x)}{g(x)}=0$$

To individuate the dominant term, recall that

• $\frac{x^a}{x^b} \to 0$ for $b>a$

• $\frac{x^a}{b^x} \to 0$ for $b>1$

• $\frac{\log x}{x^a} \to 0$ for $a>0$

and for sequences

• $\frac{a^n}{n!} \to 0$ for $a>1$

• $\frac{n!}{n^n} \to 0$

Recall that l'Hopital rule can be applied to limits which are expressed (or can be expressed) by quotient which are in the indeterminate form $\frac 0 0$ or $\frac{\pm \infty}{\pm \infty}$.

I don't think there is a specific name for the shorcut.

As indicated in the comment another way for rational expression can be the following

$$\frac{3x^2+14x-5}{x^2+x-12}=\frac{3x^2+3x-12+11x+7}{x^2+x-12}=3+\frac{11x+7}{x^2+x-12}$$

• @amWhy Something wrong or not clear? – gimusi May 11 '18 at 21:41
• But L'Hopital can be used in other situations too. – Namaste May 11 '18 at 21:42
• @amWhy are you referring to $0\cdot \infty$? – gimusi May 11 '18 at 21:43

On the other hand, it's not difficult to prove a general theorem about rational functions. Suppose you have $$f(x)=\frac{a_mx^m+a_{m-1}x^{m-1}+\dots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\dots+b_0}$$ with $a_m\ne0$ and $b_n\ne0$. Then, since it's not restrictive to assume $x>0$ when we want to compute the limit for $x\to\infty$, the numerator can be written as $$x^m\left(a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}\right)$$ and the factor in parentheses has limit $a_m$ when $x\to\infty$. Similarly for the denominator. Now $$\lim_{x\to\infty} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} =\frac{a_m}{b_n}\ne0$$ Hence there are three cases.

## First case: $m>n$

$$\lim_{x\to\infty}f(x)= \lim_{x\to\infty}x^{m-n} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} = \begin{cases} \infty & \text{if a_m/b_n>0} \\[4px] -\infty & \text{if a_m/b_n<0} \end{cases}$$ The factor $x^{m-n}$ has limit $\infty$ and the other factor is bounded.

## Second case: $m=n$

$$\lim_{x\to\infty}f(x)= \lim_{x\to\infty} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} =\frac{a_m}{b_n}$$

## Third case: $m<n$

$$\lim_{x\to\infty}f(x)= \lim_{x\to\infty}\frac{1}{x^{n-m}} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} =0$$ The factor $1/x^{n-m}$ has limit $0$ and the other factor has limit $a_m/b_n$.

This completely settles the problem and you need nothing else: your given limit is $3$ because the function is in case two. Doing umpteen times the same computations doesn't seem the best way to spend our time.

How can you remember this? The polynomial of greater degree dominates: if it is at the numerator, the limit is $\pm\infty$ (with the sign determined by $a_m/b_n$); if it is at the denominator, the limit is $0$. If the degrees are the same, the limit is $a_m/b_m$.

• This is essentially the mnemonic “Bobo Bots eats DC”—bigger on bottom: $0$; bigger on top: slant; exponents are the same: divide coefficients. – gen-z ready to perish May 11 '18 at 22:23
• @ChaseRyanTaylor I don't like such kind of mnemonics, TBH. But this is quite a simple case to handle and a general rule is surely handy. There's a similar case for integrals of the form $\int x^ne^x\,dx$; rather than doing umpteen integrations by parts, it's much easier to remember the antiderivative is of the form $P(x)e^x$ with $P$ of degree $n$ and do the derivative. – egreg May 11 '18 at 22:32

In general, suppose you have an expression of the form

$$f_1(x)+f_2(x)\over g_1(x)+g_2(x)$$

for which

$$\lim_{x\to c}{f_2(x)\over f_1(x)}=\lim_{x\to c}{g_2(x)\over g_1(x)}=0$$ Then

$$\lim_{x\to c}{f_1(x)+f_2(x)\over g_1(x)+g_2(x)}=\lim_{x\to c}{f_1(x)\over g_1(x)}$$

The proof amounts to inserting the intermediate expression

$$\lim_{x\to c}{f_1(x)\over g_1(x)}\cdot{1+{f_2(x)\over f_1(x)}\over1+{g_2(x)\over g_1(x)}}$$

A nice example, in which L'Hopital gets you nowhere, is

$$\lim_{x\to\infty}{5e^{3x}+2e^x\over2e^{3x}+7e^{2x}}=\lim_{x\to\infty}{5e^{3x}\over2e^{3x}}=\lim_{x\to\infty}{5\over2}={5\over2}$$

The trick, in general, is to recognize a dominant term in the numerator and/or denominator. In essence, the shortcut says you can ignore all the other stuff. But you have to make sure that you pick out terms that really do dominate the other stuff; sometimes you can eyeball it, and sometimes you make mistakes (at least I do).

• Have you got an explicit description of what a dominant term is? – rainier May 11 '18 at 22:44
• @rainier, it's what the second display says: the quotient of the "other stuff" to the "dominant" term tends to $0$ in the limit. In the example, we have $(2e^x)/(5e^{3x})={2\over5}e^{-2x}\to0$ and $(7e^{2x})/(2e^{3x})={7\over2}e^{-x}\to0$ as $x\to\infty$. – Barry Cipra May 11 '18 at 22:49
• Got it, thanks! – rainier May 11 '18 at 22:53