# Examples On Continuity Set-3

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Example - 15

Let f\left( x \right) = \left\{ {\begin{align}&{\frac{{\sin 2x + \sin x}}{x}} {\,\,\,\,\,\qquad \quad\; x < 0}\\& \qquad \;\;\; a\qquad\qquad\qquad\quad{x = 0}\\&{\frac{{\sqrt {x + b{x^2}} - \sqrt x }}{{b{x^{3/2}}}}} {\qquad\quad\; x > 0}\end{align}} \right\}. Find the values of  a and b so that f (x) is continuous at  x = 0.

Solution:    $${\rm{LHL}}\left( {{\rm{at }}x = 0} \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$$

\begin{align}&= \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin 2x + \sin x}}{x}\\\\& = \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{2.\sin 2x}}{{2x}} + \frac{{\sin }}{x}} \right\}\\& = {\rm{ }}3\end{align}

$$f\left( 0 \right){\rm{ }} = a$$

$${\rm{RHL}}\left( {{\rm{at}}\,x\,{\rm{ = }}\,{\rm{0}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$$

\begin{align}&= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + b{x^2}} - \sqrt x }}{{b{x^{3/2}}}}\\\\& = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + b{x^2}} - \sqrt x }}{{b{x^{3/2}}}} \times \frac{{\left( {\sqrt {x + b{x^2}} + \sqrt x } \right)}}{{\left( {\sqrt {x + b{x^2}} + \sqrt x } \right)}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{b{x^2}}}{{b{x^{3/2}}\left( {\sqrt {x + b{x^2}} + \sqrt x } \right)}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt x }}{{\sqrt x + \sqrt {x + b{x^2}} }}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + \sqrt {1 + bx} }}\\& = \frac{1}{2}\end{align}

Since  $${\rm{LHL}} \ne {\rm{RHL,}}$$ no such  values of  a  and  b  exist that  could make the  function  continuous

at x = 0

Example - 16

Let  F\left( x \right) = \left\{ {\begin{align}&{{{\left( {1 + x + \frac{{f\left( x \right)}}{x}} \right)}^{\frac{1}{x}}},} \qquad{x \ne 0}\\&{{e^3}\qquad\qquad \qquad\quad \,,\qquad} {x = 0}\end{align}} \right\} {\rm{and\,\,\, }}G\left( x \right) = \left\{ {\begin{align}{{{\left( {1 + \frac{{f\left( x \right)}}{x}} \right)}^{\frac{1}{x}}},} \quad{x \ne 0}\\{k\quad\quad\quad},\quad {x = 0}\end{align}} \right\}

where f (x) is some  function of x. If F(x) is continuous at x = 0, find the value of k so that G (x) is also continuous at x = 0.

Solution:   Since F (x) is continuous at x = 0,

\begin{align}\mathop {\lim }\limits_{x \to 0} F\left( x \right) &= F\left( 0 \right) = {e^3}\\\\ & \Rightarrow \mathop {\lim }\limits_{x \to 0} {\left( {1 + x + \frac{{f\left( x \right)}}{x}} \right)^{\frac{1}{x}}} = {e^3}\\\\ &\Rightarrow \mathop {\lim }\limits_{x \to 0} \,\,{e^{\frac{1}{x}.\left\{ {x + \frac{{f\left( x \right)}}{x}} \right\}}} = {e^3}\\\\&\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {1 + \frac{{f\left( x \right)}}{{{x^2}}}} \right) = 3\\\\ &\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{{x^2}}} = 2\end{align}

For G(x) to be continuous at x = 0, k should be equal to $$\mathop {\lim }\limits_{x \to 0} G\left( x \right)$$

\begin{align}\mathop {\lim }\limits_{x \to 0} G\left( x \right) & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + \frac{{f\left( x \right)}}{x}} \right)^{\frac{1}{x}}}\\ &= {e^{\mathop {\lim }\limits_{x \to 0} }}^{\frac{1}{x}\,.\,\left( {\frac{{f\left( x \right)}}{x}} \right)}\\\\ &= {e^{\mathop {\lim }\limits_{x \to 0} }}^{\frac{{f\left( x \right)}}{{{x^2}}}}\\& = {e^2}\end{align}

Therefore,

$$k = {{\text{e}}^{2}}$$

Example – 17

Show that the function  f\left( x \right) = \left\{ \begin{align} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}}\;\;\;{\mkern 1mu} {\rm{when}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x \ne 0\\0\quad \;\;\;{\mkern 1mu} \quad \;\;\;{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{when}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x = 0 \end{align} \right\}  is discontinuous at x = 0

Solution:    $${\rm{LHL}}\left( {{\rm{at}}\,x = 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( { - h} \right)$$

\begin{align} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - 1/h}} - 1}}{{{e^{ - 1/h}} + 1}}\end{align}

Now, as  $$h \to 0,\frac{{ - 1}}{h} \to - \infty$$  so $${e^{ - 1/h}} \to 0$$

Therefore , LHL = –1

Similarly,

$${\rm{RHL}} = \mathop {\lim }\limits_{h \to 0} f\left( h \right)$$

\begin{align}& = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{1/h}} - 1}}{{{e^{1/h}} + 1}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{1 - {e^{ - 1/h}}}}{{1 + {e^{ - 1/h}}}}\\\\ &= 1\end{align}

Since $${\rm{LHL}} \ne {\rm{RHL}} \ne f\left( 0 \right),f\left( x \right)$$   is discontinuous at x = 0

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