Question

If $f\left( x \right) = \left\{ \begin{gathered}
  \dfrac{{{e^{2x}} - 1}}{{ax}},{\text{ for }}x < 0,a \ne 0 \\
  1,{\text{ for }}x = 0 \\
  \dfrac{{\log \left( {1 + 7x} \right)}}{{bx}},{\text{ for }}x > 0,b \ne 0 \\
\end{gathered} \right\}$ is continuous at x = 0, then find the values of a and b.

Answer 1

Hint: Here, we will proceed by finding the left hand limit (LHL), the right hand limit (RHL) and the value of the function at the point where the function is given continuous (x = 0). Then, we will use the concept that LHL at that point = Value of the function at that point = RHL at that point.

Complete step-by-step answer:
Given function is $f\left( x \right) = \left\{ \begin{gathered}
  \dfrac{{{e^{2x}} - 1}}{{ax}},{\text{ for }}x < 0,a \ne 0 \\
  1,{\text{ for }}x = 0 \\
  \dfrac{{\log \left( {1 + 7x} \right)}}{{bx}},{\text{ for }}x > 0,b \ne 0 \\
\end{gathered} \right\}$
LHL at x = 0 = $f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( {x < 0} \right)$
$ \Rightarrow f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{{e^{2x}} - 1}}{{ax}}$
Put x = 0-h $ \Rightarrow $x = -h (by doing so $\mathop {\lim }\limits_{x \to {0^ - }} $ will become $\mathop {\lim }\limits_{h \to 0} $), the above equation becomes
\[
   \Rightarrow f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{2\left( { - h} \right)}} - 1}}{{a\left( { - h} \right)}} \\
   \Rightarrow f\left( {{0^ - }} \right) = - \left( {\dfrac{1}{a}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{ - 2h}} - 1}}{h} \\
 \]
The above limit corresponds to $\dfrac{0}{0}$ form which can be solved using L-Hospital’s rule in which both the numerator and denominator of the limit is differentiated with respect to the variable (i.e., h)
\[ \Rightarrow f\left( {{0^ - }} \right) = - \left( {\dfrac{1}{a}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{d}{{dh}}\left( {{e^{ - 2h}} - 1} \right)}}{{\dfrac{{dh}}{{dh}}}}\]
Using the formula \[\dfrac{d}{{dh}}\left( {{e^{ch}}} \right) = c{e^{ch}}\] (where c is any constant) in the above equation, we get
\[
   \Rightarrow f\left( {{0^ - }} \right) = - \left( {\dfrac{1}{a}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2{e^{ - 2h}}}}{1} \\
   \Rightarrow f\left( {{0^ - }} \right) = \left( {\dfrac{2}{a}} \right)\mathop {\lim }\limits_{h \to 0} {e^{ - 2h}} \\
   \Rightarrow f\left( {{0^ - }} \right) = \left( {\dfrac{2}{a}} \right)\left( {{e^{ - 2 \times 0}}} \right) \\
   \Rightarrow f\left( {{0^ - }} \right) = \left( {\dfrac{2}{a}} \right)\left( {{e^0}} \right) \\
   \Rightarrow f\left( {{0^ - }} \right) = \left( {\dfrac{2}{a}} \right)\left( 1 \right) \\
   \Rightarrow f\left( {{0^ - }} \right) = \dfrac{2}{a}{\text{ }} \to {\text{(1)}} \\
 \]
Value of the function at x = 0 = \[f\left( 0 \right) = f\left( {x = 0} \right)\]
\[ \Rightarrow f\left( 0 \right) = 1{\text{ }} \to {\text{(2)}}\]
RHL at x = 0 = $f\left( {{0^ + }} \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( {x > 0} \right)$
$ \Rightarrow f\left( {{0^ + }} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\log \left( {1 + 7x} \right)}}{{bx}}$
Put x = 0+h $ \Rightarrow $x = h (by doing so $\mathop {\lim }\limits_{x \to {0^ + }} $ will become $\mathop {\lim }\limits_{h \to 0} $), the above equation becomes
\[
   \Rightarrow f\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {1 + 7h} \right)}}{{bh}} \\
   \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{1}{b}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {1 + 7h} \right)}}{h} \\
 \]
The above limit corresponds to $\dfrac{0}{0}$ form which can be solved using L-Hospital’s rule in which both the numerator and denominator of the limit is differentiated with respect to the variable (i.e., h)
\[ \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{1}{b}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{d}{{dh}}\left( {\log \left( {1 + 7h} \right)} \right)}}{{\dfrac{{dh}}{{dh}}}}\]
Using the formula \[\dfrac{d}{{dh}}\left( {\log \left[ {f\left( h \right)} \right]} \right) = \dfrac{1}{{f\left( h \right)}}\dfrac{d}{{dh}}\left( {f\left( h \right)} \right)\] in the above equation, we get
\[
   \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{1}{b}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\dfrac{1}{{1 + 7h}}} \right)\left[ {\dfrac{d}{{dh}}\left( {7h} \right)} \right]}}{1} \\
   \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{1}{b}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{7}{{1 + 7h}} \\
   \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{7}{b}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{1}{{1 + 7h}} \\
   \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{7}{b}} \right)\left[ {\dfrac{1}{{1 + \left( {7 \times 0} \right)}}} \right] \\
   \Rightarrow f\left( {{0^ + }} \right) = \left( {\dfrac{7}{b}} \right)\left[ {\dfrac{1}{{1 + \left( {7 \times 0} \right)}}} \right] \\
   \Rightarrow f\left( {{0^ + }} \right) = \dfrac{7}{b}{\text{ }} \to {\text{(3)}} \\
 \]
Given, this function is continuous at x = 0 so the left hand limit at x = 0 will be equal to the value of the function at x = 0 which will be further equal to the right hand limit at x = 0
i.e., $f\left( {{0^ - }} \right) = f\left( 0 \right) = f\left( {{0^ + }} \right)$
By using equations (1), (2) and (3), we get
$\dfrac{2}{a} = 1 = \dfrac{7}{b}{\text{ }} \to {\text{(4)}}$
Taking $\dfrac{2}{a} = 1$ from equation (4), we have
$ \Rightarrow a = 2$
Taking $1 = \dfrac{7}{b}$ from equation (4), we have
$ \Rightarrow b = 7$
Therefore, the values of a and b are 2 and 7 respectively.

Note:In this particular problem, for the LHL of the function at x = 0, the considered function is $f\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{ax}}$ because this is the definition of the function for x<0. For the value of the function at x = 0, $f\left( x \right) = 1$ as given. For the RHL of the function at x = 0, the considered function is \[f\left( x \right) = \dfrac{{\log \left( {1 + 7x} \right)}}{{bx}}\] because this is the definition of the function for x> 0.