L’Hopital’s Rule
Hello and welcome to this video about L’Hôpital’s Rule! When taking certain types of limits, you’ll find this 300-year-old rule can come in extremely handy.
Guillaume François Antoine de l’Hôpital was a French mathematician in the late 1600s who rubbed elbows with the likes of the Bernoulli brothers and one of the fathers of calculus himself, Gottfried Leibniz. In 1696, when he was about 35 years old, he published his book Infinitesimal Calculus with Applications to Curved Lines, the first textbook on infinitesimal calculus, and the first book to contain the famous rule. Though the rule was given to him by Johann Bernoulli two years prior, stirring some controversy, it bears l’Hôpital’s name and is still in use today.
Let’s start by taking some limits and establishing where L’Hôpital’s Rule comes into play and why it is so important.
First, consider \(\lim_{x\rightarrow 5}\frac{x^2-25}{x-5}\). Substituting 5 into the expression yields the indeterminate form \(\frac{0}{0}\), but factoring the numerator results in \(\frac{(x+5)(x-5)}{(x-5)}\), which reduces to \(x+5\). Substitute 5 again to find a limit of 10.
Recall that an indeterminate form means that we cannot yet tell if a limit exists. The idea is that if we substitute the value the input approaches and the result is indeterminate, do some algebra and substitute again. If we get a limit (or there for sure isn’t one), great. If the result is indeterminate, do some more algebra and try again. Repeat as needed.
Take a look at this limit: \(\lim_{x\rightarrow ∞}\frac{5x-2x^3}{3x^3+1}\). As x approaches ∞, the function approaches \(\frac{∞}{∞}\), indeterminate once again. We are only concerned with the cubed terms in the numerator and denominator because they’ll grow the fastest and determine the overall limit. So, only considering those, the limit becomes \(\lim_{x\rightarrow ∞}\frac{-2x^3}{3x^3}\), which is \(-\frac{2}{3}\).
Unfortunately, these methods of finding limits do not always work.
Suppose we were going to evaluate this limit:
\(\lim_{x\rightarrow 0}\frac{sin x}{x}\)Substituting 0, we’ll get \(\frac{0}{0}\). No problem, right? We got an indeterminate form, so all we need to do is a little bit of algebra. But now we have a serious problem. That quotient won’t simplify algebraically. So we can’t yet tell if there’s a limit, but we have no algebraic way of finding one. Or do we?
It’s easy to see on the graph that this limit is 1 (note that the function is undefined at x=0), but of course graphs are not always practical.
L’Hopital’s Rule says we can do this: \(\lim_{x\rightarrow 0}\frac{sinx}{x}=\lim_{x\rightarrow 0}\frac{\frac{d}{dx}(sin x)}{\frac{d}{dx}(x)}=\lim_{x\rightarrow 0}\frac{cos x}{1}=cos 0=1\)
It’s not a coincidence that we find a limit of 1, just as the graph showed.
The idea is not exactly intuitive at first, but with a \(\frac{0}{0}\) limit, it’s an application of local linearity. Really, really close to x=0, sin x looks like the line tangent to sin x there.
In other words, since the derivative of sin x is cos x, the graph of sin x close to 0 looks like its tangent line at 0, which has slope cos(0), or 1. Of course, the function in the denominator, x, looks exactly the same.
Because of local linearity, the ratio of function derivatives is usually a very good approximation of the ratio of function values.
Take a look at this one. When we go to take the limit
\(\lim_{x\rightarrow ∞}\frac{e^x}{2x^3}\)we end up with the indeterminate form with no way to simplify algebraically. In the case of 00, we “zoom in” really close to where we need to look. In an case like this, we’ll use the same method, but look out at one of the ends. Since both functions approach infinity as x approaches infinity, the question is: which one goes to infinity faster?
So, using L’Hopital’s Rule:
\(\lim_{x\rightarrow ∞}\frac{ex^2}{x^2}=\lim_{x\rightarrow ∞}\frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(2x^2)}=\lim_{x\rightarrow ∞}\frac{e^x}{4x}\)Since both functions still approach infinity, apply the rule again!
\(\lim_{x\rightarrow ∞}\frac{e^x}{4x}=\lim_{x\rightarrow ∞}\frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(4x)}=\lim_{x\rightarrow ∞}\frac{e^x}{4}=∞ \)After the second application, it’s clear we have a limit of infinity.
Like many mathematical rules, L’Hopital’s Rule won’t always work 100% of the time. It is a powerful technique to learn because it will get you out of many tight spots!
Here is the rule formally stated:
If \(\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{0}{0}\) or \(\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{±∞}{±∞}\), then
\(\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\)Let’s try a few more!
\(\lim_{x\rightarrow 2}\frac{x^3-7xT2+10x}{x^2-x-2}\)On the first attempt, the result is \(\frac{0}{0}\). Time for L’Hopital’s Rule!
\(\lim_{x\rightarrow 2}\frac{x^3-7x^2+10x}{x^2-x-2}=\lim_{x\rightarrow 2}\frac{3x^2-14x+10}{2x-1}=-\frac{6}{3}=-2\)Or what about:
\(\lim_{x\rightarrow -3}\frac{sin(πx)}{x^2-9}\)The first substitution yields \(\frac{0}{0}\).
\(\lim_{x\rightarrow -3}\frac{sin(πx)}{x^2-9}=\lim_{x\rightarrow -3}\frac{πcos (πx)}{2x}=\frac{π}{6}\)Let’s try a natural log:
\(\lim_{t\rightarrow ∞}\frac{ln(5t)}{t^2}\)Both functions approach infinity as x approaches infinity. But which one approaches infinity faster?
\(\lim_{t\rightarrow ∞}\frac{ln(5t)}{t^2}=\lim_{t\rightarrow ∞}\frac{\frac{1}{t}}{2t}=\lim_{t\rightarrow ∞}\frac{1}{2t^2}=0\)So that’s L’Hopital’s Rule, a valuable addition to any limit-taking repertoire. I hope that this video helped you understand the rule, when to use it, and how it works!
Thanks for watching, and happy studying!
