L'H�pital's Rule

Discovered By

Guillaume de l'Hopital

The Rule

For $\frac {0}{0}$ or $\frac {\infty}{\infty}$ case, the limit $\lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)}$ where $f'(x)$ and $g'(x)$ are the first derivatives of $f(x)$ and $g(x)$ , respectively.

Examples

$\lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48}$

$\lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1}$

Proof of l'Hôpital's rule

A standard proof of l'Hôpital's rule uses Cauchy's mean value theorem. l'Hôpital's rule has many variations depending on whether c and L are finite or infinite, whether f and g converge to zero or infinity, and whether the limits are one-sided or two-sided.

Zero over zero

Suppose that c and L are finite and f and g converge to zero.

First, define (or redefine) f(c) = 0 and g(c) = 0. This makes f and g continuous at c, but does not change the limit (since, by definition, the limit does not depend on the value at the point c). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, there is an interval (c − δ, c + δ) such that for all x in the interval, with the possible exception of x = c, both $f'(x)$ and $g'(x)$ exist and $g'(x)$ is not zero.

If x is in the interval (c, c + δ), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [c, x] (and a similar statement holds for x in the interval (c − δ, c)). The mean value theorem implies that g(x) is not zero (since otherwise there would be a y in the interval (c, x) with $g'(y)=0$ ). Cauchy's mean value theorem now implies that there is a point ξ_x in (c, x) such that

$\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.$

If x approaches c, then ξ_x approaches c (by the squeeze theorem ). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, it follows that

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

Infinity over infinity

Suppose that L is finite, c is positive infinity, and f and g converge to positive infinity.

For every ε > 0, there is an m such that

$\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.$

The mean value theorem implies that if x > m, then g(x) ≠ g(m) (since otherwise there would be a y in the interval (m, x) with $g'(y)=0$ ). Cauchy's mean value theorem applied to the interval [m, x] now implies that

$\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.$

Since f converges to positive infinity, if x is large enough, then f(x) ≠ f(m). Write

$\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.$

Now,

$\begin{align}

& \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. \end{align}$ (Error compiling LaTeX. Unknown error_msg)

For x sufficiently large, this is less than ε and therefore

$\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.$ *

Note: Steps are missing.

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