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
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.
No, because limit of \(\tan (x)\) at \(x = \frac{ \pi }{2}\) is not defined.
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.
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
