Limits Continuity Differentiability And Differentiation Set-3

Go back to  'SOLVED EXAMPLES'

Example – 9

If   \(f(x) = \left\{ \begin{align}&\frac{{\sin 3x + a\sin 2x + b\sin x}}{{{x^5}}}, & x \ne 0\\&\qquad \qquad\quad c & , \quad x = 0 \end{align} \right\}\) is continuous at x = 0, find a, b and c.

Solution:    For continuity, we require:

\(\begin{align}&\mathop {\lim }\limits_{x \to 0} f(x) = c \quad \left\{ \begin{array}{l}{\rm{LHL\,and\,RHL\,will\,be\,the\,same\,so\,we }}\\{\rm{do\,not\,need\,to\,evaluate\,them\, separately}}\end{array} \right\}\\\\&\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\sin 3x + a\sin 2x + b\sin x}}{{{x^5}}}} \right\}\end{align}\)

\(\begin{align}& = \mathop {\lim }\limits_{x \to 0} \frac{{\left\{ {3x - \frac{{{{(3x)}^3}}}{{3!}} + \frac{{{{(3x)}^5}}}{{5!}} - ....} \right\} + a\left\{ {2x - \frac{{{{(2x)}^3}}}{{3!}} + \frac{{{{(2x)}^5}}}{{5!}} - .....} \right\} + b\left\{ {x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ...} \right\}}}{{{x^5}}}\\\end{align}\)

 As in Section -1, example-3, note that

\[3{\rm{\ }} + {\rm{ }}2a + b = {\rm{ }}0\qquad\qquad\qquad...{\rm{ }}\left( i \right)\]

\[27{\rm{ }} + {\rm{ }}8a + b = {\rm{ }}0{\rm{ }}\qquad\qquad\qquad...{\rm{ }}\left( {ii} \right)\]

\[\left\{ \begin{align}&{\rm{If\,this\,does\,not\,hold,}}\\ & \mathop {\lim }\limits_{x \to 0} \frac{{(3 + 2a + b)x}}{{{x^5}}}{\rm{ \;and\;}}\mathop {\lim }\limits_{x \to 0} \frac{{(27 + 8a + b){x^3}}}{{{x^5}}}\\&{\rm{will\,become\,infinite}}\end{align} \right\}\]

Solving (i) and (ii), we get

\(a = {\rm{ }}-{\rm{ }}4{\rm{ }}\qquad\qquad{\rm{ }}b = {\rm{ }}5\)

The limit now reduces to

\(\begin{align}&\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{243}}{{120}} + \frac{{32a}}{{120}} + \frac{b}{{120}}} \right){x^5} + ....\left( {{\rm{higher}}\,{\rm{order}}\,{\rm{terms}}} \right)}}{{{x^5}}}\\& = \frac{{243 + 32a + b}}{{120}}\\& = {\rm{ }}1\end{align}\)

Hence,

c = 1                           

Example - 10

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

Solution:    Since   \(f\left( x \right)\)  is continuous at   \(x = 0,\)

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

To evaluate  \(f\left( 0 \right),\)  we substitute \(x = y = 0\)  in the given functional relation to get.

\(\begin{align}&\quad\quad{\rm{ }}f\left( 0 \right) = f{\left( 0 \right)^2}\\ &\Rightarrow f\left( 0 \right) = 0\,\,{\rm{or}}\,\,\,f\left( 0 \right) = 1\end{align}\)

(i)    If \(f\left( 0 \right) = 0,\)  then \(f\left( x \right) = f\left( {x + 0} \right) = f\left( x \right) \cdot f\left( 0 \right) = 0\)   i.e

\(f\left( x \right) = 0\)    for all values of x so that \(f\left( x \right)\)   is continuous everywhere.

(ii)      We now assume \(f\left( 0 \right) = 1.\)

Now,

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

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

Similarly,

                      RHL at any \(x = \mathop {\lim }\limits_{h \to 0} f\left( {x + h} \right)\)

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

Hence, \(f\left( x \right)\)   is continuous for all     \(x \in \mathbb{R}\)    

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum