L'Hôpital's Rule

Suppose f and g are continuous at a. Consider the limit


lim
x→ a 
   f(x)
g(x)
 .
If g(a) ≠ 0, then

lim
x→ a 
   f(x)
g(x)
 = f(a)
g(a)
 .

Example       limx→ 3 (x2+1)/(x+2) = 10/5 = 2.

But what happens if both the numerator and the denominator tend to 0? It is not clear what the limit is. In fact, depending on what functions f(x) and g(x) are, the limit can be anything at all!

Example

limx→ 0 (x3)/(x2) = limx→ 0 x = 0.
             limx→ 0 (-x)/( x3) = limx→ 0 -1/( x2) = -∞.
limx→ 0 x/( x2) = limx→ 0 1/x = ∞.
             limx→ 0 kx/x = limx→ 0 k = k.

These limits are examples of indeterminate forms of type 0/0. L'Hôpital's Rule provides a method for evaluating such limits. We will denote limx→ a, limx→ a+, limx→ a-, limx→ ∞, and limx→ -∞ generically by lim in what follows.

L'Hôpital's Rule for 0/0

Suppose lim f(x) = lim g(x) = 0. Then

  1. If lim (f ′(x))/( g ′(x)) = L, then lim f(x)/ g(x) = lim(f ′(x))/( g ′(x)) = L.
  2. If lim (f ′(x))/( g ′(x)) tends to +∞ or -∞ in the limit, then so does f(x)/ g(x).

Geometrical Interpretation     Sketch of the Proof of L'Hôpital's Rule

Examples

  • limx→ 0 sinx / x = limx→0 (d/dx(sinx))/(d/dx(x)) = limx→ 0 cosx / 1 = 1.
  • limx→ 1 2lnx/(x-1) = limx→1 (d/dx(2lnx))/(d/dx(x-1)) = limx→ 1 ( 2/x )/1 = 2.
  • limx→ 0 (ex-1)/( x2) = limx→0 (d/dx(ex-1))/d/dx(x2)) = limx→ 0 (ex)/2x = ∞.

If the numerator and the denominator both tend to ∞ or -∞, L'Hôpital's Rule still applies.

L'Hôpital's Rule for ∞ / ∞

Suppose lim f(x) and lim g(x) are both infinite. Then

  1. If lim (f ′(x))/( g ′(x)) = L, then lim f(x)/g(x) = lim(f ′(x))/( g ′(x)) = L.
  2. If lim (f ′(x))/( g ′(x)) tends to +∞ or -∞ in the limit, then so does f(x)/g(x).

The proof of this form of L'Hôpital's Rule requires more advanced analysis.

Here are some examples of indeterminate forms of type ∞ / ∞.

Example       limx→∞(ex)/x = limx→∞ (ex)/1 = ∞.

Sometimes it is necessary to use L'Hôpital's Rule several times in the same problem.

Example       limx→ 0 (1-cosx)/(x2) = limx→ 0sinx/ 2x = limx→ 0cosx/ 2 = 1/2.

Occasionally, a limit can be re-written in order to apply L'Hôpital's Rule.

Example       limx→ 0 x lnx = limx→ 0lnx/( 1/x) = limx→ 0 ( 1/x )/( -1/( x2)) = limx→ 0 (-x) = 0.

We can use other tricks to apply L'Hôpital's Rule. In the next example, we use L'Hôpital's Rule to evaluate an indeterminate form of type 00.

Example

To evaluate limx→ 0+ xx, we will first evaluate limx→ 0+ ln(xx).


lim
x→ 0+ 
  ln(xx) =
lim
x→ 0+ 
  xln(x) = 0,   by the previous example.
Then since limx→ 0+ ln(xx)→ 0 as x→0+ and ln(u) = 0 if and only if u = 1,
xx→ 1    as    x→ 0+.
Thus,

lim
x→ 0+ 
  xx = 1.
Notice that L'Hôpital's Rule only applies to indeterminate forms. For the limit in the first example of this tutorial, L'Hôspital's Rule does not apply and would give an incorrect result of 6. L'Hôpital's Rule is powerful and remarkably easy to use to evaluate indeterminate forms of type 0/0 and ∞ / ∞.

Key Concepts [index]

L'Hôpital's Rule for 0/0

Suppose lim f(x) = lim g(x) = 0. Then

  1. If lim (f ′(x))/ (g ′(x)) = L, then lim f(x)/g(x) = lim(f ′(x))/ (g ′(x)) = L.
  2. If lim (f ′(x))/ ( g ′(x)) tends to +∞ or -∞ in the limit, then so does f(x)/g(x).

L'Hôpital's Rule for ∞ / ∞

Suppose lim f(x) and lim g(x) are both infinite. Then

  1. If lim (f ′(x))/ ( g ′(x)) = L, then lim f(x)/g(x) = lim(f ′(x))/ (g ′(x)) = L.
  2. If lim (f ′(x))/(g ′(x)) tends to +∞ or -∞ in the limit, then so does f(x)/g(x).