Examples On Continuity Set-2

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Example – 12

Find the condition on f (x) and g (x) which makes the function

\(\begin{align}F\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right) + {x^{2n}}g\left( x \right)}}{{1 + {x^{2n}}}}\end{align}\) continuous everywhere.

Solution: Note from the definition of F (x) that we have a variable n present. Lets first try to make F (x) independent of  n.

The obvious way is to consider three separate cases for x:

\(\left| x \right| = 1,\left| x \right| > 1,\left| x \right| < 1\)

\(\begin{align}\boxed{\left| x \right| = 1}:{\text{  }}F\left( x \right) = \frac{{f\left( x \right) + \left( 1 \right) \cdot g\left( x \right)}}{{1 + 1}} = \frac{{f\left( x \right) + g\left( x \right)}}{2}\end{align}\)

\(\begin{align}\boxed{\left| x \right| > 1}:{\text{  }}F\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right) + {x^{2n}}g\left( x \right)}}{{1 + {x^{2n}}}}\end{align}\)

\(\begin{align}& = \mathop {\lim }\limits_{n \to \infty } \frac{{{x^{ - 2n}}f\left( x \right) + g\left( x \right)}}{{{x^{ - 2n}} + 1}}\\\end{align}\)

\( = g(x)\left( {{\rm{because\;}}{x^{ - 2n}} \to 0{\mkern 1mu} {\mkern 1mu} {\rm{as}}{\mkern 1mu} {\mkern 1mu} n \to \infty {\mkern 1mu} {\mkern 1mu} {\rm{for}}{\mkern 1mu} {\mkern 1mu} \left| x \right| > 1} \right)\)

\(\begin{align}\boxed{\left| x \right| < 1}:{\text{  }}F\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right) + {x^{2n}}g\left( x \right)}}{{1 + {x^{2n}}}}\end{align}\)

\( = f(x)\left( {{\rm{because\;}}{x^{2n}} \to 0{\mkern 1mu} {\mkern 1mu} {\rm{as}}{\mkern 1mu} {\mkern 1mu} n \to \infty {\mkern 1mu} {\mkern 1mu} {\rm{for}}{\mkern 1mu} {\mkern 1mu} \left| x \right| < 1} \right)\)

Therefore, we can now rewrite F(x) independently of n  in the following manner

\[F\left( x \right) = \left\{ {\begin{align}&{f\left( x \right)} {{\rm{\qquad\qquad\; when}}\,\,\,\left| x \right| < 1} {\,\,\,\,\;{\rm{or}}\,\,\,\,\, - 1 < x < 1{\rm{ }}}\\\\&{\frac{{f\left( x \right) + g\left( x \right)}}{2}} {{\rm{\quad when }}\,\,\,\left| x \right| = 1} {{\,\,\,\,\rm{or}}\,\,\,\,\,\,x = 1,\, - 1}\\\\&{g\left( x \right)} {{\rm{\quad\qquad \qquad when}}\left| x \right| > 1} {{\,\,\,\,\;\rm{or }}\,\,\,\,\,x < - 1,x > 1} \end{align}} \right\}\]

 F (x) could be discontinuous at only 2 points, x  = 1 or  x  =  –1

To ensure continuity at these points

\[{\rm{LHL }}\left( {{\rm{at }}x =  - 1} \right) = {\rm{RHL}}\left( {{\rm{at }}x =  - 1} \right) = F\left( { - 1} \right)\]

\[\begin{align}&g\left( { - 1} \right) = f\left( { - 1} \right) = \frac{{f\left( { - 1} \right) + g\left( { - 1} \right)}}{2}\\\\& \Rightarrow  g\left( { - 1} \right) = f\left( { - 1} \right)\end{align}\]

\[\begin{align}&{\rm{LHL }}\left( {{\rm{at }}x = 1} \right) = {\rm{RHL}}\left( {{\rm{at }}x = 1} \right) = F\left( 1 \right)\\\\&f\left( 1 \right) = g\left( 1 \right) = \frac{{f\left( 1 \right) + g\left( 1 \right)}}{2}\\\\&\Rightarrow  f\left( 1 \right) = g\left( 1 \right)\end{align}\]

Therefore, for continuity of F (x),

\(\begin{align}&{f\left( 1 \right){\rm{ }} = g\left( 1 \right)}\\&{f\left( {-1} \right){\rm{ }} = g\left( {-1} \right)}
\end{align}\)

Example – 13

If \(f\left( x \right) = \left\{ {\begin{align}&{{{\left( {1 + \left| {\sin x} \right|} \right)}^{\frac{a}{{\left| {\sin x} \right|}}}},} {\frac{{ - \pi }}{6} < x < 0}\\&{b,}\quad {x = 0}\\&{{e^{\frac{{\tan 2x}}{{\tan 3x}}}},} {0 < x < \frac{\pi }{6}}\end{align}} \right\}\)  is continuous at x = 0, find the values of a and b.

Solution: For continuity at x = 0,

\(\begin{align}LHL{\rm{ }}(at\;x = {\rm{ \;}}0){\rm{ \;}}&= f\left( 0 \right){\rm{ }} = {\rm{ }}RHL{\rm{ }}(at\;x = {\rm{ }}0)\\{\rm{LHL}}\left( {{\rm{at }}x = 0} \right)&= \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\\
& = \mathop {\lim }\limits_{x \to {0^ - }} {\left( {1 + \left| {\sin x} \right|} \right)^{\frac{a}{{\left| {\sin x} \right|}}}}\\ &= \mathop {\lim }\limits_{y \to 0} {\left( {1 + y} \right)^{\frac{1}{y}.a}}\left\{ \begin{array}{l}{\rm{where }}\left| {\sin x} \right| = y;\\{\rm{as }}x \to {0^ - },y \to 0\end{array} \right\}\\& = {e^a}\end{align}\)

\(\begin{align}{\rm{RHL}}\left( {{\rm{at}}\,x = 0} \right) &= \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\\& = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\frac{{\tan 2x}}{{\tan 3x}}}}\\& = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\frac{2}{3}.\frac{{\tan 2x}}{{2x}}\frac{{3x}}{{\tan 3x}}}}\\&= {e^{2/3}}\\ &\Rightarrow {e^a} = b = {e^{2/3}}\\& \Rightarrow a = \frac{2}{3}{\rm{\,\, and \,\,}}b = {e^{2/3}}\end{align}\)

Example - 14

Let \(f:\mathbb{R} \to \mathbb{R}\)  be a function satisfying \(f\left( {x + y} \right) = f\left( x \right) + f\left( y \right){\text{ for all }}x,y\, \in \mathbb{R}\)   . If the function f (x) is continuous at x = 0 show that it is continuous for all \(x\, \in \mathbb{R}\)

Solution: In questions that have such functional equations, we should try to substitute certain trial values for the variables to gain useful information; in this particular question, since we are given some condition about f (x) at x = 0, we should try to find f (0)

Putting x = y = 0, we get

\(\begin{align}&f(0) = 2f(0)\\&\Rightarrow f(0) = 0\end{align}\)

Now, since f (x) is continuous at x = 0,

\(\begin{align}&LHL{\rm{ }} = {\rm{ }}RHL{\rm{ }} = {\rm{ }}f{\rm{ }}\left( 0 \right)\\&\Rightarrow \qquad \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( 0 \right) = 0\end{align}\)

or equivalently

\(\mathop {\lim }\limits_{h \to 0} f\left( { - h} \right) = \mathop {\lim }\limits_{h \to 0} f\left( h \right) = f\left( 0 \right) = 0\)

For continuity at an arbitrary value of x, the LHL, RHL and f(x) should be equal.

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

\(\begin{align}& = \mathop {\lim }\limits_{h \to 0} \left\{ {f\left( x \right) + f\left( { - h} \right)} \right\}\\\\ &= \mathop {\lim }\limits_{h \to 0} f\left( x \right) + \mathop {\lim }\limits_{h \to 0} f\left( { - h} \right)\\\\ &= f\left( x \right)\end{align}\)

Similarly,

\(RHL{\rm{ }}\left( {at\;x} \right){\rm{ }} = f\left( x \right)\)

Therefore, f (x) is continuous for all values of x

 

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