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## Saturday, 2 August 2014

### CHAPTER 7- L’Hospital’s Rule

The $LH$ rule can be used to evaluate limits that are of the form $\dfrac{0}{0}\,\,{\rm{or}}\,\,\dfrac{\infty }{\infty }$.
Consider two functions $f\left( x \right)\,$ 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 $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
provided that the limit $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ exists. Although the $LH$ rule is applicable to limits of the form $\dfrac{0}{0}\,\,{\rm{or}}\,\,\dfrac{\infty }{\infty }$, 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 algebtain manipulations.
You are urged to think of some (non-rigorous) justification for this rule.
Lets apply this rule on some examples.
 Example: 1
Evaluate:
 (a) $\mathop {\lim }\limits_{x \to 0} x\ln x$ (b) $\mathop {\lim }\limits_{x \to \infty } \,\,\dfrac{{\ln x}}{x}$
 Solution: 1
We have encountered both these limits in the unit on Limits. Here, we’ll re-evaluate them using the $LH$ rule.
 $L = \mathop {\lim }\limits_{x \to 0} x\ln x\,\,\,\, \left( {0\,\, \times - \infty \,\,{\rm{form}}} \right)$ $= \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{1/x}}\,\,\,\,\left( {\dfrac{{ - \infty }}{\infty }\,\,{\rm{form}}} \right)$ $= \mathop {\lim }\limits_{x \to 0} \dfrac{{1/x}}{{ - 1/{x^2}}}\,\,\,\,{\rm{By\, applying\, the\, LH\, rule}}$ $= \mathop {\lim }\limits_{x \to 0} \left( { - x} \right)$ $=0$
This is what we got earlier
 Solution: 1
 $L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln x}}{x} \,\,\,\, \left( {\dfrac{\infty }{\infty }\,\,\,{\rm{form}}} \right)$ $\,\,\,\, = \mathop {\lim }\limits_{x \to \infty } \dfrac{{1/x}}{1}\,\,\,\,{\rm{By\, applying\, the\, LH\, rule}}$ $=0$
 Example: 2
 Evaluate $\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\dfrac{1}{x}} \right)^{\sin x}}$
 Solution: 2
This limit is of the indeterminate form ${\infty ^0}$. Lets first convert it into the form $\dfrac{0}{0}\,\,{\rm{or}}\,\,\dfrac{\infty }{\infty }$.
 $L = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {\dfrac{1}{x}} \right)^{\sin x}}$ $= \mathop {\lim }\limits_{x \to {0^ + }} \sin x \cdot \ln \left( {\dfrac{1}{x}} \right) \,\, \left\{ \begin{array}{l} {\rm{Doing \,this\, is\, justified}}\,\,{\rm{since \, for \,}}\,\\ x > 0\,\ln \left( {\dfrac{1}{x}} \right){\rm{is\, defined}} \end{array} \right\}$ $= - \mathop {\lim }\limits_{x \to {0^ + }} \sin x \cdot \ln x$ $= - \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\ln x}}{{{\rm{cosec}}\,x}}\,\,\,\,\left( {\dfrac{\infty }{\infty }\,\,{\rm{form}}} \right)$ $= - \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{1/x}}{{ - \cos ecx\cot x}}\,\,\,\,{\rm{By\, applying\, the\,LH\,rule}}$ $= \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{{\sin }^2}x}}{{x\cos x}}$ $= \mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{{\sin x}}{x}} \right) \cdot \tan x$ $=0$
 Example 3
 Evaluate $\mathop {\lim }\limits_{x \to 1} \left( {\dfrac{1}{{\ln x}} - \dfrac{x}{{\ln x}}} \right)$
 Solution: 3
 $L = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{1 - x}}{{\ln x}}} \right)\,\,\,\,\left( {\dfrac{0}{0}{\rm{ form}}} \right)$ $= \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{ - 1}}{{1/x}}} \right)\,\,\,\,\left( \begin{array}{l} {\rm{By \,applying\, the\, }}\\ {\rm{LH \,rule\,}} \end{array} \right)$ $= \mathop {\lim }\limits_{x \to 1} \left( { - x} \right)$ $=1$
 Example: 4
 Evaluate $\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + \sin x}}{{{x^2}}}} \right)$
 Solution: 4
The limit is of the indeterminate form $\dfrac{\infty }{\infty }$, so we apply the $L.H.$ rule:
 $L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + \sin x}}{{{x^2}}}} \right)\,\,\,\,\left( {\dfrac{\infty }{\infty }{\rm{ form}}} \right)$ $= \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2x + \cos x}}{{2x}}} \right)\,\,\,\,\left( {\dfrac{\infty }{\infty }{\rm{ form \,again}}} \right)$ $(1)$ $= \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2 - \sin x}}{2}} \right)$ $(2)$ $= 1 - \dfrac{1}{2}\mathop {\lim }\limits_{x \to \infty } \,\,(\sin x)$
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 $\left( {\dfrac{{{x^2} + \sin x}}{{{x^2}}}} \right)$ carefully:
 $\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + \sin x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{{\sin x}}{{{x^2}}}} \right)$ $= 1 + \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\sin x}}{{{x^2}}}} \right)$ $= 1 + 0$ $\left( {{\rm{because}}\,\sin x\,\,{\rm{is\, bounded}}} \right)$ $= 1$
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 $\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$ 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.