L’Hôpital’s rule

L’Hôpital’s rule (pronounced “low-pee-tal”) can sometimes be used to work out limits of quotients.

Definition

When we work out the limit of a function at a point, we can sometimes get help from derivatives. Suppose we have a function in the form of a quotient, that is, we have functions f(x) and g(x) and we are looking for the limit

$$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}.$$

If $\underset{x\to x_0}{lim}f(x)=\underset{x\to x_0}{lim}g(x)=0$, both functions have a derivative in some neighbourhood of the point x₀ (they need not have one at x₀ itself) and the limit

$$\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)},$$

exists, then

$$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}.$$

If the limit $\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ does not exist, that does not mean the limit $\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}$ does not exist either. If the limit with derivatives does not exist, we have to work out the original limit in some other way, without derivatives.

Examples

1. Work out the limit

   $$\underset{x\to 3}{lim}\frac{x^2-9}{x^2-x-6}$$

   Can we solve this example using L’Hôpital’s rule? First we have to check that the limit of both functions (the one in the denominator of the fraction and the one in the numerator) at the point x₀ = 3 is zero. Let’s call the functions f(x) = x² − 9 and g(x) = x² − x − 6. Both are polynomials and therefore continuous, so their limits are equal to their values:

   $$\begin{array}{rcl}f(3)& =& 3^2-9=0\\ g(3)& =& 3^2-3-6=0\end{array}$$

   The first condition is satisfied. Notice that we cannot work out the limit simply by substituting three for x, because the whole function is discontinuous at x₀ = 3 – in fact it has no value there at all (to get one we would have to divide by zero). Now we can try L’Hôpital’s rule and differentiate both functions. Watch out: we do not differentiate using the quotient rule – we differentiate the function in the numerator and the function in the denominator separately. So we get:

   $$\begin{array}{rcl}{f}^{\prime }(x)& =& 2x\\ g^{\prime }(x)& =& 2x-1\end{array}$$

   Instead of the previous limit, we can now work out this one:

   $$\underset{x\to 3}{lim}\frac{2x}{2x-1}$$

   This function is continuous at x₀ = 3, so we can simply substitute three for x and work out the value of the whole function:

   $$\underset{x\to 3}{lim}\frac{2x}{2x-1}=\frac{2\cdot 3}{2\cdot 3-1}=\frac{6}{5}$$

   We can see that the limit of this function exists, so the limit $\underset{x\to 3}{lim}\frac{x^2-9}{x^2-x-6}$ exists too and the two limits are equal. So:

   $$\underset{x\to 3}{lim}\frac{x^2-9}{x^2-x-6}=\underset{x\to 3}{lim}\frac{2x}{2x-1}=\frac{6}{5}.$$

2. Work out the limit

   $$\underset{x\to 0}{lim}\frac{x}{\sin x}$$

   What is the situation at x₀ = 0? Again let f(x) = x and g(x) = sin x. We have f(0) = 0 and g(0) = 0. We get the quotient $\frac{0}{0}$ and the function is discontinuous at that point. We can try L’Hôpital’s rule. Differentiate both functions:

   $$\begin{array}{rcl}{f}^{\prime }(x)& =& 1\\ g^{\prime }(x)& =& \cos x\end{array}$$

   Now that we have the derivatives, let’s try to work out the limit

   $$\underset{x\to 0}{lim}\frac{1}{\cos x}$$

   What happens at x₀ = 0? Since cos(0) = 1, we get the fraction $\frac{1}{1}=1$. The function is continuous at x₀ = 0, so

   $$\underset{x\to 0}{lim}\frac{1}{\cos x}=\frac{1}{1}=1.$$

   Since this limit exists, the limit $\underset{x\to 0}{lim}\frac{x}{\sin x}$ exists too and the two limits are equal. We get

   $$\underset{x\to 0}{lim}\frac{x}{\sin x}=1$$