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     

(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)
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}}
     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
     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)
     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.
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