Limits Continuity Differentiability And Differentiation Set-2

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Example – 5

\(\mathop {\lim }\limits_{n \to \infty } \frac{{48}}{{{n^4}}}\left[ {1\left( {\sum\limits_{k = 1}^n k } \right) + 2\left( {\sum\limits_{k = 1}^{n - 1} k } \right) + 3\left( {\sum\limits_{k = 1}^{n - 2} k } \right) + ... + n.1} \right] = \)

 

(a) 0           (b) 1                      (c) 2                             (d) none of these

Solution:    The general rth  term in the series is  \({t_r} = r\left( {\sum\limits_{k = 1}^{n - r + 1} k } \right)\) which equals \(r \cdot \frac{{\left( {n - r + 1} \right)\left( {n - r + 2} \right)}}{2}.\)

Thus,

\(\sum\limits_{r = 1}^n {{t_r} = \sum\limits_{r = 1}^n {\frac{{r\left( {n - r + 1} \right)\left( {n - r + 2} \right)}}{2}} } \)

\(\begin{align}& = \sum\limits_{s = 0}^{n - 1} {\frac{{\left( {s + 1} \right)\left( {n - s} \right)\left( {n - s + 1} \right)}}{2}} \\&\ = \sum\limits_{s = 0}^{n - 1} {\frac{1}{2}\left( {{s^3} - 2n{s^2} + \left( {{n^2} - n - 1} \right)s + {n^2} + n} \right)} \end{align}\)

The highest degree term in this summation is  \(\frac{{{n^4}}}{{24}}.\)

Thus, the limit is \(\frac{{48}}{{{n^4}}} \times \frac{{{n^4}}}{{24}} = 2.\)

Example – 6

Let   \(f\left( {\frac{{x + y}}{3}} \right) = \frac{{f\left( x \right) + f\left( y \right) + 2}}{3} \;\;\;\forall x,y \in \mathbb{R}.\,f'\left( 0 \right) = 2.\)

(a)  \(f'\left( x \right) = f'\left( 0 \right)\;\;\;\;\forall x \in \mathbb{R}\)   (b)  \(f'\left( x \right) = f\left( x \right)\;\;\;\;\forall x \in \mathbb{R}\)
(c)  \(f\left( x \right)\)  is a linear function  (d)  none of these

Solution: \(f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\)

\(\begin{align}& = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{f\left( {3x} \right) + f\left( {3h} \right) + 2}}{3} - \frac{{f\left( {3x} \right) + f\left( 0 \right) + 2}}{3}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {3h} \right) - f\left( 0 \right)}}{{3h}}\\ &= f'\left( 0 \right) = 2\\\ &\Rightarrow  \quad f\left( x \right) = 2x + c{\rm{\;\; }}which{\rm{\; }}is{\rm{\; }}a{\rm{ \;}}linear{\rm{ }}function \end{align}\)

Example - 7

Evaluate   \(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sin x - {{\left( {\sin x} \right)}^{\sin x}}}}{{1 - \sin x + \ln \left( {\sin x} \right)}}\)

Solution:   \(L = \mathop {\lim }\limits_{t \to 1} \frac{{t - {t^t}}}{{1 - t + \ln t}}{\rm{              }}\left( {t = \sin x} \right)\)

\(\begin{align} & \qquad =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( 1+h \right)-{{\left( 1+h \right)}^{1+h}}}{-h+\ln \left( 1+h \right)} \\ 
&\qquad =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( 1+h \right)\left\{ 1-{{\left( 1+h \right)}^{h}} \right\}}{-h+\left\{ h-\frac{{{h}^{2}}}{2}+\frac{{{h}^{3}}}{3}... \right\}} \\ & \qquad =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( 1+h \right)\left\{ 1-\left( 1+{{h}^{2}}+h\frac{\left( h-1 \right){{h}^{2}}}{2!}+... \right) \right\}}{-\ \frac{{{h}^{2}}}{2}\ +\ \frac{{{h}^{3}}}{3}\ -...} \\&\qquad=\text{ }2 \\ \end{align}\)

Example – 8

Evaluate   \(\mathop {\lim }\limits_{k \to \infty } \left\{ {\frac{{{e^{1/k}} + 2{e^{2/k}} + 3{e^{3/k}} + ...k{e^{k/k}}}}{{{k^2}}}} \right\}\)

Solution:    The numerator is an A.G.P. which sums to

\[s = \frac{{{e^{1/k}}\left( {e - 1} \right)}}{{{{\left( {{e^{1/k}} - 1} \right)}^2}}} + \frac{{k{e^{1 + 1/k}}}}{{{e^{1/k}} - 1}}\]

Now, \(\mathop {\lim }\limits_{k \to \infty } \frac{s}{{{k^2}}}\)   can be reduced to 1.  Since all limits in this expression take the form  \(\mathop {\lim }\limits_{y \to 0} \left( {\frac{{{e^y} - 1}}{y}} \right).\)

Therefore, the required limit is 1.

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