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Saturday, 9 August 2014

chapter 13 Worked Out Examples 4

   Example: 6    

Evaluate the following limits:
(a) \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - \cos \,x}}{{{x^2}}}
(b) \mathop {\lim }\limits_{x \to 0} \dfrac{{x - \ln \left( {1 + x} \right)}}{{{x^2}}}
(c) \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^{\dfrac{1}{x}}} - e}}{x}
(d) \mathop {\lim }\limits_{x \to 0} \dfrac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2  - \sqrt {1 + \cos \,x} }}
Solution: 6

The limit is of the indeterminate form \dfrac{0}{0}, but can be reduced into a combination of two standard limits as follows:
\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - 1 + 1 - \cos \,x}}{{{x^2}}}
 = \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{{e^{{x^2}}} - 1}}{{{x^2}}} + \dfrac{{1 - \cos \,x}}{{{x^2}}}} \right\}
 = \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{{e^{{x^2}}} - 1}}{{{x^2}}} + \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{4{{\left( {\dfrac{x}{2}} \right)}^2}}}} \right\}
 = 1 + \dfrac{1}{2} = \dfrac{3}{2}
Solution: 6-(b)

\ln \left( {1 + x} \right) can be expanded as x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \ldots
Hence,

\mathop {\lim }\limits_{x \to 0} \dfrac{{x - \ln \left( {1 + x} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} +\ldots}}{{{x^2}}}} \right\}
 = \dfrac{1}{2}
Solution: 6-(c)

The numerator in this limits tends to 0 as x \to {\rm{0}} because \mathop {\lim }\limits_{x \to 0} {(1 + x)^{\dfrac{1}{x}}}e
Evaluating this limit will require a little artifice in the following manner:
\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{{{(1 + x)}^{\dfrac{1}{x}}} - e}}{x}} \right\} = \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{{e^{\dfrac{{\ln (1 + x}}{x}}} - e}}{x}} \right\}
 = e\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{{e^{\dfrac{{\ln \left( {1 + x} \right)}}{x}}} - 1}}{x}} \right\} Taking “e” common of the numerator
Now as x \to {\rm{0,}}\left\{ {\dfrac{{{\rm{ln}}\left( {{\rm{1 + }}x} \right)}}{x} - 1} \right\} \to 0 so that the numerator in the limit above is of the form {e^h} - 1 where h \to 0.
What should we do now? Multiply and divide by h\left( {h = \dfrac{{\ln \left( {1 + x} \right)}}{x} - 1} \right).
We get

From the previous example, it follows that the second limit has the value
  - \dfrac{1}{2}.
 = e\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{{e^{\dfrac{{\ln \left( {1 + x} \right)}}{x} - 1}} - 1}}{{\left\{ {\dfrac{{\ln \left( {1 + x} \right)}}{x} - 1} \right\}}}} \right] \cdot \left\{ {\dfrac{{\ln \left( {1 + x} \right)}}{x} - 1} \right\} \cdot \dfrac{1}{x}
 = e\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{{e^h} - 1}}{h}} \right] \cdot \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{\ln \left( {1 + x} \right) - x}}{{{x^2}}}} \right\}
Hence, the overall value for this limit is  - \dfrac{e}{2}
Solution: 6-(d)

Factoring the numerator and rationalising the denominator gives.
\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{({9^x} - 1)({3^x} - 1)}}{{\sqrt 2  - \sqrt {1 + \cos x} }}.\dfrac{{\sqrt 2  + \sqrt {1 + \cos x} }}{{\sqrt 2  + \sqrt {1 + \cos x} }}} \right\}
 = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{9^x} - 1} \right)}}{x} \cdot \dfrac{{\left( {{3^x} - 1} \right)}}{x} \cdot \dfrac{{\left( {\sqrt 2  + \sqrt {1 + \cos x} } \right)}}{{1 - \cos x}} \cdot {x^2}
(We have multiplied and divided by {x^2} above)
 = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{9^x} - 1} \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{3^x} - 1} \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \left( {\sqrt 2  + \sqrt {1 + \cos x} } \right) \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{2}{{\dfrac{{{{\sin }^2}x/2}}{{{{\left( {x/2} \right)}^2}}}}}
\ln 9.\ln 3.2\sqrt 2 .2
 = 8\sqrt 2 {\left( {\ln 3} \right)^2}
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