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Saturday, 9 August 2014

chapter 6 Some Standard Limits 2

(C) \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln (1 + x)}}{x} = 1\left( {\ln y \equiv {{\log }_e}y} \right)
This limit is of the indeterminate form \dfrac{0}{0} . We can easily evaluate this limit based on the previous limit.
\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x}
 = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}\ln \left( {1 + x} \right)
 = \mathop {\lim }\limits_{x \to 0} \ln {\left( {1 + x} \right)^{\dfrac{1}{x}}}
(Property of log)
 = \ln \left\{ {\mathop {\lim }\limits_{x \to 0} {{\left( {1 + x} \right)}^{\dfrac{1}{x}}}} \right\}
 = \ln e = {\log _e}e = 1
This limit can alternatively be evaluated by using the expansion series for \ln \left( {1 + x} \right)
\ln \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \ldots\infty   \,\,\,\,\,{\rm{when}}\,\left| x \right| < 1
\ln \left( {1 + x} \right) = 1 - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \ldots\infty
 \Rightarrow   \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x} = 1 (All other terms involving x tend to 0
This is just an obvious extension of the previous limit(D) \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\log }_a}\left( {1 + x} \right)}}{x} = \dfrac{1}{{\ln a}}
\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\log }_a}\left( {1 + x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{{\left( {\ln a} \right) \cdot x}}
 = \dfrac{1}{{\ln a}}\left( {\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x}} \right)
 = \dfrac{1}{{\ln a}}
We have used the following property of logarithms above: {\log _a}y = \dfrac{{{{\log }_b}y}}{{{{\log }_b}a}}
(E) {\mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x} = \ln \,\,a}
This is again an extension of the limits seen previously. Let {a^x} - 1 = t. This gives x = {\log _a}(1 + t) As x \to 0, {a^x} \to 1 and {a^x} - 1 = t \to 0
Hence, we have the limit L as
L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \dfrac{t}{{{{\log }_a}\left( {1 + t} \right)}}
 = \mathop {\lim }\limits_{t \to 0} \dfrac{1}{{\left( {\dfrac{{{{\log }_a}\left( {1 + t} \right)}}{t}} \right)}} = \dfrac{1}{{\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{{\log }_a}\left( {1 + t} \right)}}{t}} \right)}}
 = \dfrac{1}{{1/\ln a}} = \ln \,\,a.
Note that for a = e, this limit is 1.
(F) \mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x} = \ln a
When m is an integer, it is easy to see that the above relation holds because {x^m} - 1 can be expanded as (x - 1)({x^{m - 1}} + {x^{m - 2}} + \ldots + 1)

Now we have \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - 1}}{{x - 1}} = \mathop {\lim }\limits_{y \to 0} \dfrac{{{{\left( {1 + y} \right)}^m} - 1}}{y} . Now expand {\left( {1 + y} \right)^m} using the Binomial theorem for a general index.For the general case, let x = 1 + y. As x \to 1,y \to 0.
{\left( {1 + y} \right)^m} = 1 + my + \dfrac{{m\left( {m - 1} \right){y^2}}}{{2!}} + \ldots
Hence,
\mathop {\lim }\limits_{y \to 0} \dfrac{{{{\left( {1 + y} \right)}^m} - 1}}{y}
 = \mathop {\lim }\limits_{y \to 0} \dfrac{{my + \dfrac{{m\left( {m - 1} \right)}}{{2!}}{y^2} + \ldots}}{y}
 = \mathop {\lim }\limits_{y \to 0} \left( {m + \dfrac{{m\left( {m - 1} \right)}}{{2!}}y + \ldots} \right)
=m (all other terms tend to 0
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