L-Hospital Rule

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We had made a mention of the L'Hospital's Rule (abbreviated as the LH rule) in the unit on limits. We had deferred the introduction of this rule since it requires the use of differentiation. The LH rule can be used to evaluate limits that are of the form \(\frac{0}{0}\,\,{\rm{or}}\,\,\frac{\infty }{\infty }\) .

Consider two functions \(f\left( x \right)\,\,{\rm{and}}\,\,g\left( x \right)\) which are differentiable in the neighbourhood of the point x = a (except possibly at the point x = a itself). Let \(g'\left( x \right) \ne 0\) in this neighbourhood.

The LH rule says that

if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} g\left( x \right) = 0\)

or if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} g\left( x \right) = \infty \)

then \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)

provided that the limit \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}}\) exists. Although the LH rule is applicable to limits of the form \(\frac{0}{0}\,\,{\rm{or}}\,\,\frac{\infty }{\infty }\) , you should be able to understand that other indeterminate forms like \(0.\infty ,\,\,\infty - \infty ,\,\,{1^\infty },\,\,{\infty ^0}\,\,{\rm{or}}\,\,\,{0^0}\) can be reduced to these two indeterminate forms using appropriate algebraic manipulations.

You are urged to think of some (non-rigorous) justification for this rule.

Lets apply this rule on some examples.

Example – 32


(a) \(\mathop {\lim }\limits_{x \to 0} x\ln x\)    (b) \(\mathop {\lim }\limits_{x \to \infty } \,\,\frac{{\ln x}}{x}\)

Solution: We have encountered both these limits in the unit on Limits. Here, we’ll re-evaluate them using the LH rule.

(a)                       \(L = \mathop {\lim }\limits_{x \to 0} x\ln x \left( {0\,\, \times - \infty \,\,{\rm{form}}} \right)\)

\(\begin{align}& =\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln x}{1/x}\text{ }\left( \frac{-\infty }{\infty }\,\,\text{form} \right) \\& =\underset{x\to 0}{\mathop{\lim }}\,\frac{1/x}{-1/{{x}^{2}}}~~~~~~~~~~~\left( By\text{ }applying\text{ }the\text{ }LH\text{ }rule \right) \\ &=\underset{x\to 0}{\mathop{\lim }}\,\left( -x \right) \\ &=\text{ }0 \\ \end{align}\)

This is what we got earlier

(b)                    \(L = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{x} \left( {\frac{\infty }{\infty }\,\,\,{\rm{form}}} \right)\)

\(\begin{align} &=\underset{x\to \infty }{\mathop{\lim }}\,\frac{1/x}{1}\text{ }\left( By\text{ }applying\text{ }the\text{ }LH\text{ }rule \right) \\ &=\text{ }0 \\ \end{align}\)

Example – 33

Evaluate \(\begin{align}\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{1}{x}} \right)^{\sin x}}\end{align}\)

Solution: This limit is of the indeterminate form \({\infty ^0}.\) Lets first convert it into the form \(\frac{0}{0}\,\,{\rm{or}}\,\,\frac{\infty }{\infty }\) .

\(L = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{1}{x}} \right)^{\sin x}}\)

    \( = \mathop {\lim }\limits_{x \to {0^ + }} \sin x \cdot \ln \left( {\frac{1}{x}} \right)  \left\{ \begin{align}&{\rm{Doing\;this\;is\;justified\;}}\,\,{\rm{since\;}}\,\\&x > 0\ln \left( {\frac{1}{x}} \right){\rm{is\;defined}}\end{align} \right\}\)

\(\begin{align}& = - \mathop {\lim }\limits_{x \to {0^ + }} \sin x \cdot \ln x\\ &= - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{{\rm{cosec}}\,x}} & \left( {\frac{\infty }{\infty }\,\,{\rm{form}}} \right)\end{align}\)

  \( = - \mathop {\lim }\limits_{x \to {0^ + }} {\mkern 1mu} \frac{{\frac{1}{x}\;}}{{ - {\rm{cosec}}\cot x}}\)    (By applying the LH rule)

\(\begin{align}& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\sin }^2}x}}{{x\cos x}}\\ &= \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\sin x}}{x}} \right) \cdot \tan x\\& = 0\end{align}\)

Example - 34

Evaluate \(\mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{{\ln x}} - \frac{x}{{\ln x}}} \right)\)

Solution: \(L = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 - x}}{{\ln x}}} \right){\rm{ }}\left( {\frac{0}{0}{\rm{ form}}} \right)\)

\(\begin{align}& = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{ - 1}}{{1/x}}} \right){\rm{ }}\left( \begin{array}{l}{\rm{By\;applying\;the\;}}\\{\rm{LH \;rule}}\end{array} \right)\\& = \mathop {\lim }\limits_{x \to 1} \left( { - x} \right)\\&= - 1\end{align}\)

Example - 35

Evaluate \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + \sin x}}{{{x^2}}}} \right)\)

Solution: The limit is of the indeterminate form \(\frac{\infty }{\infty }\) , so we apply the L.H. rule:

\(L = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + \sin x}}{{{x^2}}}} \right){\rm{ }}\left( {\frac{\infty }{\infty }{\rm{form}}} \right)\)

\(\begin{align}& = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2x + \cos x}}{{2x}}} \right){\rm{ }}\left( {\frac{\infty }{\infty }{\rm{ form\;again}}} \right){\rm{ }}...\left( 1 \right)\\& = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2 - \sin x}}{2}} \right){\rm{ }}...\left( 2 \right)\\&= 1 - \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \,\,(\sin x) \end{align}\)

Now, we know that \(\mathop {\lim }\limits_{x \to \infty } (\sin x)\) does not exist since sin x is an oscillating function and does not converge to any particular value. What does this imply for our current limit?

Does it not exist ?

Think about the expression \(\begin{align}\left( {\frac{{{x^2} + \sin x}}{{{x^2}}}} \right)\end{align}\) carefully:

\[\begin{align}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + \sin x}}{{{x^2}}}} \right) &= \mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{{\sin x}}{{{x^2}}}} \right)\\&= 1 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\sin x}}{{{x^2}}}} \right)\\& = 1 + 0 \left( {∵\sin x\,\,{\rm{is bounded}}} \right)\\& = 1\end{align}\]

Thus, a limit does infact exists while the LH rule says that it does not exist. Why?

This is because the LH rule is not applicable here. Go back to the definition of the LH rule which says that \(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'(x)}}{{g'(x)}}\) if the latter limit exists.

In this example, you cannot apply the LH rule on the expression in (1) since the limit for the expression obtained after differentiation (the one in (2)) does not exist.

Thus, the LH rule must be used with care.


1. Differentiate the following functions:

(a) \(f\left( x \right) = \cos \sqrt {1 + {x^3}} \)    (b) \(f\left( x \right) = {e^{{{\tan }^{ - 1}}\sqrt x }}\)  (c) \(f\left( x \right) = {\left( {\sin x} \right)^{\tan x}}\)
(d) \(f\left( x \right) = {\tan ^2}\left( {{x^3} + \sin \sqrt x } \right)\)  (e) \(f\left( x \right) = \frac{{\sin x}}{{1 + {e^x}}}\)    (f) \(f\left( x \right) = \sin \left( {\cos \left( {\tan \left( {{e^x}} \right)} \right)} \right)\)

 2. Use the LH rule to evaluate \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}\)

3. If \(u = f\left( {{x^2}} \right),v = g\left( {{x^3}} \right),f'\left( x \right) = \sin x\;{\rm{and}}\;g'\left( x \right) = \cos x,\) find \(\frac{{du}}{{dv}}\)

4. Let f be twice differentiable such that \(f''\left( x \right)=-f\left( x \right)\ \text{and}\ f'\left( x \right)=g\left( x \right).\) If \(h\left( x \right) = {\left( {f\left( x \right)} \right)^2} + {\left( {g\left( x \right)} \right)^2}\) where \(h\left( 5 \right) = 11,\;{\rm{find}}\;h\left( {10} \right).\)

5. Use this series

\[\cos x + \cos 3x + \cos 5x + ... + \cos \left( {2k - 1} \right)x = \frac{{\sin 2kx}}{{2\sin x}}\]

to evaluate the series

\[S = \sin x + 3\sin 3x + 5\sin 5x + ..... + \left( {2k - 1} \right)\sin \left( {2k - 1} \right)x\]

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