Indeterminate Forms and l'Hospital's Rule

Consider $x \to 1 lim \frac{ln x}{x - 1}$ . Both numerator and denominator approach $0$ , and $\frac{0}{0}$ is not a number, so the limit laws for quotients do not apply: the value is not determined by the form. It might be any number, or fail to exist. The same ambiguity arises for $x \to \infty lim \frac{e ^{x}}{x ^{2}}$ , where numerator and denominator both grow without bound. l'Hospital's Rule handles both cases by replacing the ratio of two functions with the ratio of their derivatives, and applied repeatedly it ranks the competing growth rates of logarithms, powers, and exponentials.

Indeterminate forms

Some such limits yield to algebra. A rational $\frac{0}{0}$ cancels a common factor,

x \to 1 lim \frac{x ^{2} - x}{x ^{2} - 1} = x \to 1 lim \frac{x ( x - 1 )}{( x + 1 ) ( x - 1 )} = x \to 1 lim \frac{x}{x + 1} = \frac{1}{2},

and $x \to 0 lim \frac{sin x}{x} = 1$ was found by a geometric argument. But $\frac{ln x}{x - 1}$ has no factor to cancel, and a general method is needed.

l'Hospital's Rule

The statement holds for one-sided limits and for $a = \pm \infty$ ; the symbol $x \to a$ may be replaced by $x \to a^{+}$ , $x \to a^{-}$ , $x \to \infty$ , or $x \to - \infty$ . Named for the Marquis de l'Hospital, who published it in 1696, the rule was discovered by John Bernoulli.

The reason is visible when $f (a) = g (a) = 0$ with $f^{'}, g^{'}$ continuous and $g^{'} (a) \neq = 0$ . Near $a$ each function is close to its tangent line through $(a, 0)$ , so the ratio of the functions is close to the ratio of the two linear approximations, which is the ratio of their slopes.

When both functions vanish at a, zooming in makes each nearly linear; the ratio of the functions approaches the ratio of the tangent slopes f'/g'.

Formally, using the definition of the derivative,

x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{\frac{f ( x ) - f ( a )}{x - a}}{\frac{g ( x ) - g ( a )}{x - a}} = \frac{f ^{'} ( a )}{g ^{'} ( a )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )},

where the middle step uses $f (a) = g (a) = 0$ . The general proof, valid without continuity of the derivatives and for the $\frac{\infty}{\infty}$ case, is more delicate but gives the same conclusion.

The differentiation is of numerator and denominator separately; this is not the quotient rule. Returning to the opening example,

x \to 1 lim \frac{ln x}{x - 1} = x \to 1 lim \frac{1/ x}{1} = 1.

Comparing growth rates

Repeated application compares the growth rates of the function families of this module. For $x \to \infty lim \frac{e ^{x}}{x ^{2}}$ , type $\frac{\infty}{\infty}$ , two passes clear the denominator:

x \to \infty lim \frac{e ^{x}}{x ^{2}} = x \to \infty lim \frac{e ^{x}}{2 x} = x \to \infty lim \frac{e ^{x}}{2} = \infty.

The exponential outgrows the power; in fact $e^{x}$ grows faster than every power $x^{n}$ . At the other extreme the logarithm grows slower than every positive power:

x \to \infty lim \frac{ln x}{3 x} = x \to \infty lim \frac{1/ x}{\frac{1}{3} x ^{- 2/3}} = x \to \infty lim \frac{3}{x ^{1/3}} = 0.

These two results give the ordering that governs limits at infinity: logarithms grow slower than powers, and powers grow slower than exponentials.

The growth race at large x. The logarithm trails every power, and every power trails the exponential, so ln x, x^2, and e^x separate in order.

A harder $\frac{0}{0}$ example needs three applications:

x \to 0 lim \frac{tan x - x}{x ^{3}} = x \to 0 lim \frac{sec ^{2} x - 1}{3 x ^{2}} = x \to 0 lim \frac{2 sec ^{2} x tan x}{6 x} = \frac{1}{3} x \to 0 lim \frac{tan x}{x} = \frac{1}{3} .

Checking the form

l'Hospital's Rule applies only to $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . Applying it blindly to a ratio that is not indeterminate gives a wrong answer. For $x \to π^{-} lim \frac{sin x}{1 - cos x}$ , the numerator tends to $0$ but the denominator tends to $1 - (- 1) = 2 \neq = 0$ , so the form is $\frac{0}{2}$ , not indeterminate, and the function is continuous there:

x \to π^{-} lim \frac{sin x}{1 - cos x} = \frac{sin π}{1 - cos π} = \frac{0}{2} = 0.

Differentiating top and bottom would have produced $cos x / sin x \to - \infty$ , which is incorrect. Verify the form before differentiating, and prefer simpler methods when they apply.

A second failure mode is a genuine indeterminate form on which the rule cycles without ever simplifying. There the derivatives only reshuffle the same expression, and elementary algebra settles the limit at once.

A limit on which l'Hospital cycles. Factoring x from the radical shows the ratio decreasing to the horizontal asymptote y = 1.

Reducing other forms to a quotient

Products, differences, and powers can each be indeterminate, and each is handled by algebra that recasts it as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

Form	Situation	Conversion
$0 \cdot \infty$	$f \to 0$ , $g \to \infty$	write $f g = \frac{f}{1/ g}$ or $\frac{g}{1/ f}$
$\infty - \infty$	$f \to \infty$ , $g \to \infty$	common denominator, factor, or rationalize
$0^{0}, \infty^{0}, 1^{\infty}$	$f^{g}$	take $ln$ : $ln y = g ln f$ , type $0 \cdot \infty$

Indeterminate products

For type $0 \cdot \infty$ , move one factor into the denominator.

Indeterminate differences

For type $\infty - \infty$ , combine into a single fraction.

Indeterminate powers

Three forms arise from $y = [f (x)]^{g (x)}$ : type $0^{0}$ when $f \to 0$ and $g \to 0$ ; type $\infty^{0}$ when $f \to \infty$ and $g \to 0$ ; and type $1^{\infty}$ when $f \to 1$ and $g \to \pm \infty$ . All three are resolved by taking the logarithm, which turns the power into the product $g ln f$ .

A power indeterminacy is resolved through the logarithm: take ln to get a product, convert to a quotient, apply the rule, then exponentiate.

For $x \to 0^{+} lim (1 + sin 4 x)^{c o t x}$ , type $1^{\infty}$ , set $y = (1 + sin 4 x)^{c o t x}$ , so

ln y = cot x ln (1 + sin 4 x) = \frac{ln ( 1 + sin 4 x )}{tan x},

a $\frac{0}{0}$ form. l'Hospital gives

x \to 0^{+} lim ln y = x \to 0^{+} lim \frac{\frac{4 cos 4 x}{1 + sin 4 x}}{sec ^{2} x} = 4,

and since $y = e^{ln y}$ , the original limit is $e^{4}$ . The exponentiation at the end is essential: the rule computes $lim ln y$ , and the answer is $e$ raised to that.

The canonical $1^{\infty}$ form is the limit that defines $e$ .

The function x^x for x > 0 descends from the limit 1 at the origin to a minimum near x = 1/e, then rises past 1.

The type $\infty^{0}$ is handled identically. For $x \to \infty lim x^{1/ x}$ , set $y = x^{1/ x}$ , so $ln y = \frac{ln x}{x}$ , a $\frac{\infty}{\infty}$ form. l'Hospital gives

x \to \infty lim ln y = x \to \infty lim \frac{1/ x}{1} = 0, so x \to \infty lim x^{1/ x} = e^{0} = 1.

The vanishing exponent cancels the growing base, leaving $1$ .

Proof of the rule

The special case with $f (a) = g (a) = 0$ and continuous derivatives was proved above by splitting the ratio into two difference quotients. The general statement, which allows one-sided approach and the $\frac{\infty}{\infty}$ form, rests on a two-function version of the Mean Value Theorem.

Taking $g (x) = x$ recovers the ordinary Mean Value Theorem. Applied to the pair $(f, g)$ on a shrinking interval with endpoint $a$ , and using $f (a) = g (a) = 0$ , the equation rearranges to

\frac{f ( x )}{g ( x )} = \frac{f ( x ) - f ( a )}{g ( x ) - g ( a )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}

for some $c$ between $a$ and $x$ . As $x \to a$ the intermediate point $c$ is squeezed to $a$ as well, so the ratio $f^{'} (c) / g^{'} (c)$ approaches $lim f^{'} (x) / g^{'} (x)$ , which is the conclusion of the rule.