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Limits Continuity Differentiability And Differentiation Set-1

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Example - 1

(a) Find \(\begin{align}\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{a_1^{1/x} + a_2^{1/x} + ... + a_n^{1/x}}}{n}} \right)^{nx}}\end{align}\) where all the \({a_i}'s\)  are positive.

(b)  Evaluate \(\begin{align}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\log }_{\sin x}}\cos x}}{{{{\log }_{{{\sin }_{x/2}}}}\cos \left( {x/2} \right)}}\end{align}\)

Solution:    (a) The given limit L is of the form \({1^\infty }\) . Thus,

\[\begin{align}L &= {e^{\mathop {\lim }\limits_{x \to \infty } \,nx\;\left\{ {\;\frac{{a_1^{1/x} + a_2^{1/x} + ...a_n^{1/x}}}{n} - 1\;} \right\}}}\\&= {e^{\mathop {\lim }\limits_{x \to \infty } \,\;\left\{ {\;\frac{{\left( {a_1^{1/x} - 1} \right) + \left( {a_2^{1/x} - 1} \right) + ... + \left( {a_n^{1/x} - 1} \right)}}{{1/x}}\;} \right\}}}\\ &= {e^{\left( {\ln {a_1} + \ln {a_2} + ... + \ln {a_n}} \right)}} = {a_1}\;{a_2}\;{a_3}\;...{a_n}\end{align}\]

(b) The given limit L is of the form \(\frac{0}{0}\) :

\[\begin{align}L &= \mathop {\lim }\limits_{x \to {0^ + }} \;\frac{{\ln \cos x}}{{\ln \cos x{\rm{/}}2}}\; \cdot \;\;\frac{{\ln \sin x{\rm{/}}2}}{{\ln \sin x}}\\&= \mathop {\lim }\limits_{x \to {0^ + }} \;\frac{{ - 2\tan x}}{{\tan x{\rm{/}}2}}\; \cdot \;\;\frac{{\cot x/2}}{{2\cot x}}\left\{\text{Using the rule on the two limits separately} \right\}\\& = - \mathop {\lim }\limits_{x \to {0^ + }} \;\frac{{{{\tan }^2}x}}{{{{\tan }^2}\frac{x}{2}}} = - \;4\end{align}\]

Example – 2

Find a and b such that \(\begin{align}\mathop {\lim }\limits_{x \to 0} \;\frac{{a{{\sin }^2}x + b\log \cos x}}{{{x^4}}} = \frac{1}{2}\end{align}\)

Solution:    Using the LH rule,

\[\begin{align}&\mathop {\lim }\limits_{x \to 0} \;\frac{{a\sin 2x - b\tan x}}{{4{x^3}}} = \frac{1}{2}\\\Rightarrow \qquad & \mathop {\lim }\limits_{x \to 0} \;\frac{{a\left( {2x - \frac{{{{(2x)}^3}}}{{3!}} + \frac{{{{(2x)}^5}}}{{5!}} - .....} \right) - b\left( {x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + .....} \right)}}{{4{x^3}}} = \frac{1}{2}\\  \Rightarrow \qquad&  a = - 1,\;b = - 2\end{align}\]

Example – 3

Let \(\begin{align}f\left( x \right) = \frac{1}{2}\left( {x + \frac{a}{x}} \right),a \ne 0.\end{align}\) If \(\mathop {\lim }\limits_{x \to \sqrt a } \frac{{f\left( x \right) - \sqrt a }}{{{{\left( {x - \sqrt a } \right)}^p}}} = m\left( { \ne 0} \right),\)  find p and m.

Solution:   \(\mathop {\lim }\limits_{x \to \sqrt a } \;\frac{{f(x) - \sqrt a }}{{{{(x - \sqrt a )}^p}}} = m\)

Applying the LH rule twice on this yields

\[\begin{align}&\mathop {\lim }\limits_{x \to \sqrt a } \;\frac{{\frac{a}{{{x^3}}}}}{{p(p - 1){{(x - \sqrt a )}^{p - 2}}}} = m\\ &\Rightarrow \qquad p = 2,\;p(p - 1) = \frac{1}{{m\sqrt a }} & \Rightarrow & m = \frac{1}{{2\sqrt a }}\end{align}\]

Example – 4

\(\mathop {\lim }\limits_{n \to \infty } \;{n^{ - {n^2}}}{\left( {\left( {n + 1} \right)\left( {n + \frac{1}{2}} \right)\left( {n + \frac{1}{{{2^2}}}} \right)...\left( {n + \frac{1}{{{2^{n - 1}}}}} \right)} \right)^n} = \)

 

(a) 1                             (b) e                             (c) \({e^2}\)                            (d) none of these

Solution:   The limit L can be written as

\[\begin{align} L & = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + \frac{1}{2}} \right)\left( {n + \frac{1}{{{2^2}}}} \right)....\left( {n + \frac{1}{{{2^{n - 1}}}}} \right)}}{{{n^n}}}} \right)^n}\\ & = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n}{\left( {1 + \frac{1}{{2n}}} \right)^n}{\left( {1 + \frac{1}{{{2^2}n}}} \right)^n}...{\left( {1 + \frac{1}{{{2^{n - 1}}n}}} \right)^n}\\ & = e\;.\;{e^{{\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle 2$}}}}\;.\;{e^{{\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {{2^2}}$}}}}\;...\;{e^{{\raise0.5ex\hbox{$\scriptstyle 1$}\kern-0.1em/\kern-0.15em\lower0.25ex\hbox{$\scriptstyle {{2^{n - 1}}}$}}}}...\infty \\\\ & = {e^2}\end{align}\]

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