**L’Hospital’s Rule** or **L’Hôpital’s Rule** (pronunciation lopeetaal rule), is a rule employed in calculation of limits of indeterminate forms \((\frac{0}{0} \mbox{ or } \frac{\infty}{\infty})\).

It is named after the French mathematician Guillaume de l’hospital (1661-1704). The essential message of l’hospital’s rule is that, under certain conditions,

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

where, \(f(x)\) represents the function in the numerator and \(g(x)\) represents the function in the denominator.

Index

**History**

The L’Hospital’s rule first made its appearance in *Guillaume de L’Hopital’s* book *Analyse des Infiniment Petits pour l’Intelligence des Lignes Courbes* (literal translation: Analysis of the Infinitely Small for the Understanding of Curved Lines) in 1696.

Although some believe that *Johann Bernoulli*, l’hospital’s academic advisor, was the one who introduced l’hospital to the rule, the rule is named after l’hospital.

**Formal Statement of L’Hospital’s Rule**

Let \(f(x)\) and \(g(x)\) be two functions differentiable on an open interval \((a, b)\) and not necessarily differentiable at a single point \(c \in (a, b)\) and

\(\lim_{x \to c } f(x) = \lim_{x \to c} g(x) = 0 \mbox{ or } \pm \infty\),

then \(\lim_{x \to c } \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\),

provided, the limit exists.

The limit of a fraction of functions is equal to the limit of the fraction of derivatives of those functions. This rule works whenever the fraction is in indeterminate form of \(\frac{0}{0} \mbox{ or } \frac{\infty}{\infty}\).

**Example.** Find \(\lim_{x \to 0} \frac{2 sin(x) – sin(2x) }{x-sin(x)}\)

**Solution.** Here \(c = 0, f(x) = 2 sin(x) – sin(2x), g(x) = x – sin(x)\)

Consider, \(\lim_{x \to 0} f(x) = \lim_{x \to 0} (2 sin(x) – sin(2x)) = 2×sin(0) – sin(0) = 0\)

Also, \(\lim_{x \to 0} g(x) = \lim_{x \to 0} (x – sin(x)) = 0 – sin(0) = 0\)

Therefore, the limit given in the question is in indeterminate form \(\frac{0}{0}\). Also \(f(x)\) and \(g(x)\) are differentiable for all \(x \in \mathbb{R}\).

Applying the L’Hospital’s Rule,

\(\lim_{x \to 0} \frac{2 sin(x) – sin(2x) }{x-sin(x)}\)

= \(\lim_{x \to 0} \frac{(2 sin(x) – sin(2x))’}{(x-sin(x))’}\)

= \(\lim_{x \to 0} \frac{2 cos(x) – 2 cos(2x)} { 1- cos(x)}\) (still in \(\frac{0}{0}\) form)

Applying the rule again,

\(\lim_{x \to 0} \frac{2 cos(x) – 2 cos(2x)} { 1- cos(x)}\)

= \(\lim_{x \to 0} \frac{ – 2 sin(x) + 4 sin(2x)} { sin(x)}\) (still in \(\frac{0}{0}\) form)

Applying the rule yet again,

\(\lim_{x \to 0} \frac{ – 2 sin(x) + 4 sin(2x)} {sin(x)}\)

= \(\lim_{x \to 0} \frac{ – 2 cos(x) + 8 cos(2x)} { cos(x)}\) (simple form)

= \(\frac{-2 +8}{1} = 6\)

**Geometrical Meaning of L’Hospital’s Rule**

Consider a 2D parametric curve defined by,

\(x = g(t)\) and \(y = f(t)\) where \(t\) is the parameter. Each point on the curve has coordinate \((g(t), f(t))\). Slope of tangent to the curve at a point \(t=c\) is

Slope of tangent at \(c = \frac{df}{dg}_t=c\)

= \(\frac{ \frac{ df}{ dt}|_{t=c}}{ \frac{ dg}{ dt}|_{t=c} }\)

= \(\frac{f'(c)}{g'(c)}\)

Consider a point on the curve where \(f(c) = g(c) = 0\) , i.e., the origin. Let \(t = c\) at this point. The equation of the tangent to this point would have the form \(y = mx\) (because tangent passes through the origin).

So, the slope of the tangent is given by,

\(m=\frac {y}{x} ; x \neq 0\)

or more generally,

\(m = \lim_{\mbox{ some } t} \frac{f(t)}{g(t)}\) *(Parametric Curve (g(t), f(t)))*

Taking \(t=c\),

slope = \(\lim_{t \to c} \frac{f(t)}{g(t)}\)

But we already know that slope at this point = \(\frac{f'(c)}{g'(c)}\), which equivalent to \(\lim_{t \to c} \frac{f'(c)}{g'(c)}\).

Therefore, \(\lim_{t \to c} \frac{f(t)}{g(t)} = \lim_{t \to c} \frac{f'(c)}{g'(c)}\)

**Proof of L’Hospital’s Rule**

Now, we attempt to prove L’Hospital’s Rule for the case when \(c\) the functions \(f\) and \(g\) are differentiable at \(c\) and \(c\) is a real number. A more general proof( at \(c\) not differentiable) is complicated and beyond our scope of study.

Let the two functions \(f(x)\) and \(g(x)\) be differentiable at \(x = c\) where \(c \in\) the interval where the functions are defined. Also let \(f(c) = g(c) = 0\).

Consider, \(\lim_{x \to c } \frac{f(x)}{g(x)}\)

= \(\lim_{x \to c } \frac{f(x) – 0 }{g(x) – 0 }\)

= \(\lim_{x \to c } \frac{f(x) – f(c) }{g(x) – g(c) }\)

= \(\lim_{x \to c } \frac{\frac {f(x) – f(c) }{ x – c }}{\frac{g(x) – g(c)}{ x- c }}\)

= \(\lim_{x \to c } \frac{f'(x) }{g'(x) }\) (By definition of a derivative)

**Examples**

**Question 1.** Find the value of \(\lim_{ x \to \infty } x^n . e^{-x}\)

**Solution.** We have, \(\lim_{ x \to \infty }x^n . e^{-x}\)

\(\lim_{ x \to \infty } \frac{x^n}{e^x}\)

Applying L’Hospital’s Rule,

\(\lim_{ x \to \infty } \frac{x^n}{e^x}\)

\(= n . \lim_{ x \to \infty } \frac{x^{n-1}}{e^x}\)

Continuously using the rule:

\(n . \lim_{ x \to \infty } \frac{x^{n-1}}{e^x}\)

= \(n! . \lim_{ x \to \infty } \frac{1}{e^x}\)

= \(n! . \lim_{ x \to \infty } \frac{0}{e^x} = 0\)

**Question 2.** Find the value of \(\lim_{x \to \infty }(1+\frac{3}{x})^x\).

**Solution.** Let, \(A = \lim_{x \to \infty }(1+\frac{3}{x})^x\)

Taking log on both sides,

\(\ln(A) = \lim_{x \to \infty } x.\ln(1+ \frac{3}{x})\)

= \(\lim_{x \to \infty } 3.\frac{\ln(1+ \frac{3}{x})}{\frac{3}{x}}\)

Let \(y = \frac{3}{x}\), so that, as \(x \to \infty\), \(y \to 0\).

Therefore, \(\ln(A) = \lim_{y \to 0 } 3. \frac{\ln(1+ y )}{ y }\)

Applying L’Hospital’s rule,

\(ln(A) = 3.\lim_{y \to 0 } \frac{1}{1+ y } = 3\)

Therefore, \(A = e^3\).

**Question 3.** Find the value of \(\lim_{x \to 0 } \frac{e^x – 1 }{x}\).

**Solution.** Applying the L’Hospital’s Rule,

\(\lim_{x \to 0 } \frac{e^x – 1 }{x}\)

= \(\lim_{x \to 0 } \frac{ e^x}{1}\)

= \(e^0 = 1\)

**Applications**

- L’Hospital’s Rule is employed daily to
**compute the indeterminate limits**of the form \(\frac{0}{0} \mbox{ or } \frac{ \infty } { \infty }\). - L’Hospital’s Rule is employed in evaluating
**compound interest formulae**in banking. - L’Hospital’s Rule is a basis for other advanced rules such as
**Stolz – Cesàro formula**.

**FAQs**

**What is L’Hospital’s Rule?**L’Hospital’s rule provides a technique to evaluate the limits of indeterminate forms. Application of the rule often converts an indeterminate form to an expression that can be easily evaluated by substitution.

**Can L’Hospital’s Rule be applied to \(\lim_{x \to \frac{ \pi } {2}} \frac{sin(x)}{cos (x) }\)?**No, because limit of \(\tan (x)\) at \(x = \frac{ \pi }{2}\) is not defined.

**When should you use L’Hopital’s Rule?**L’Hospital’s Rule is used whenever the limit is in \(\frac{0}{0}\) or \(\frac{ \infty }{ \infty }\) indeterminate form. After using the rule once, if the limit is still in indeterminate form, then use the rule again and again till the limit simplifies into a determinate form.

**How to solve limits of the form \(0^\infty\)?**First convert into \(\frac{0}{0}\) form:

Let \(A = \lim \infty^0\).

Take log on both sides,

\(ln(A) = \lim \infty×ln(0)\)

\(ln(A) = \lim \frac{ ln(0)}{ \frac{1}{ \infty}}\)

\(ln(A) = \lim \frac{0}{0}\)

Now use L’Hospital’s Rule.

**A Video On L Hospital’s Rule**