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Saturday, 9 August 2014

chapter 9 Methods for Evaluation of Limits 2

 (C) RATIONALIZATION
In this method, the rationalization of an indeterminate expression leads to determinate one. The following examples elaborate this method.
(i) \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1}  - 1}}{{\sqrt {{x^2} + 16}  - 4}}\left( {{\rm{of\, \,the\, \,indeterminate\,\, form \,\,}}\dfrac{{\rm{0}}}{{\rm{0}}}} \right)
Rationalizing both the numerator and the denominator
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1}  - 1}}{{\sqrt {{x^2} + 16}  - 4}} \times \dfrac{{\sqrt {{x^2} + 1}  + 1}}{{\sqrt {{x^2} + 1}  + 4}} \times \dfrac{{\sqrt {{x^2} + 16}  + 4}}{{\sqrt {{x^2} + 16}  + 4}}
\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{{x^2}}} \times \dfrac{{\sqrt {{x^2} + 16}  + 4}}{1} A determinate form
Cancelling out {x^2} from the numerator and the denominator
 = \dfrac{8}{2} = 4
(ii) \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1}  - \sqrt {{x^2} + 1} } \right)  \left( {{\rm{of\, the \,indeterminate\, form \,}}\infty  - \infty } \right)
 = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1}  - \sqrt {{x^2} + 1} } \right) \times \dfrac{{\sqrt {{x^2} + x + 1}  + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + x + 1}  + \sqrt {{x^2} + 1} }}
 = \mathop {\lim }\limits_{x \to \infty } \dfrac{{({x^2} + x + 1) - ({x^2} + 1)}}{{\sqrt {{x^2} + x + 1}  + \sqrt {{x^2} + 1} }}
 = \mathop {\lim }\limits_{x \to \infty } \dfrac{x}{{\sqrt {{x^2} + x + 1}  + \sqrt {{x^2} + 1} }}
\mathop {\lim }\limits_{x \to \infty } \dfrac{x}{{\sqrt {{x^2} + x + 1}  + \sqrt {{x^2} + 1} }}
Divide the numerator and denominator by x
\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}  + \sqrt {1 + \dfrac{1}{{{x^2}}}} }}
Rationalization has led us to another indeterminate form of \dfrac{\infty }{\infty }. However, it can easily be made determinate in the following manner:
Now as x \to \infty \dfrac{1}{x} \to 0 and \dfrac{1}{{{x^2}}} \to 0
Hence, the limit above reduces to
\dfrac{1}{{\sqrt {1 + 0 + 0}  + \sqrt {1 + 0} }} = \dfrac{1}{2}
(D) REDUCTION TO STANDARD FORMS
In this method, we try to reduce the given limit to one of the standard forms we studied earlier.
(i) \mathop {\lim }\limits_{x \to 0} {\left( {1 + \sin x} \right)^{2\cot x}}
This limit is of the indeterminate form{1^\infty }\left( {{\rm{as}}\,x \to 0,\sin x \to 0{\rm{and}}\cot x \to \infty } \right).
We proceed as follows
\mathop {\lim }\limits_{x \to 0} {(1 + \sin x)^{2\cot x}} = \mathop {\lim }\limits_{x \to 0}   {(1 + \sin x)^{{{\dfrac{1}{{\sin x}}}^{^{2\cos x}}}}}
{{{(1 + \sin x)}^{\frac{1}{{\sin x}}}}} This limit has the value e
= {e^{\mathop {\lim }\limits_{x \to 0} 2\cos x}} = {e^2}(\cos x \to 1\,\,\,{\rm{as}}\,\,\,x \to 0)
(ii) \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\cot x - \cos x}}{{{{(\pi  - 2x)}^3}}}
This limit is of the indeterminate form \dfrac{0}{0}

\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\cot x - \cos x}}{{{{(\pi  - 2x)}^3}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cot \left( {\dfrac{\pi }{2} + h} \right) - \cos \left( {\dfrac{\pi }{2} + h} \right)}}{{{{( - 2h)}^3}}}
 = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{ - \tan h + \sin h}}{{ - 8{h^3}}}} \right)
 = \dfrac{1}{8}\mathop {\lim }\limits_{h \to 0} \dfrac{{ - \sin h + \dfrac{{\sin h}}{{\cos h}}}}{{{h^3}}}
=\dfrac{1}{8}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin h}}{h}.\dfrac{{1 - \cos h}}{{{h^2}}}.\dfrac{1}{{\cos h}}
 = \dfrac{1}{8}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h}.\dfrac{{2{{\sin }^2}\dfrac{h}{2}}}{{4{{\left( {\dfrac{h}{2}} \right)}^2}}}.\dfrac{1}{{\cosh }}
 = \dfrac{1}{{16}}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h}.{\left( {\dfrac{{\sin \dfrac{h}{2}}}{{\left( {\dfrac{h}{2}} \right)}}} \right)^2}.\dfrac{1}{{\cosh }}
Let x = \dfrac{\pi }{2} + h so that as x \to \dfrac{\pi }{2},h \to 0.
This expression now only contains the limits \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1 and \mathop {\lim }\limits_{x \to 0} \cos x = 1
Hence, the final result is \dfrac{1}{{16}}
We will now see examples based on the methods discussed above. We urge you to first try out all these examples on your own before viewing the solutions.
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