ONE MORE STEP TOWARD BETTER TOMORROW : chapter 6 Some Standard Limits 2

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 $0$

Saturday, 9 August 2014

chapter 6 Some Standard Limits 2

No comments:

https://www.youtube.com/TarunGehlot