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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)
 $= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{9^x} - 1} \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{3^x} - 1} \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right) \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{2}{{\dfrac{{{{\sin }^2}x/2}}{{{{\left( {x/2} \right)}^2}}}}}$ $\ln 9.\ln 3.2\sqrt 2 .2$ $= 8\sqrt 2 {\left( {\ln 3} \right)^2}$