Question

Discuss the continuity of the following function at $x = 0$. If the function has a removable discontinuity, redefine the function so as to remove the discontinuity.
f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{{4^x} - {e^x}}}{{{6^x} - 1}}\,for\,x \ne 0} \\\ {\log \left( {\dfrac{2}{3}} \right)\,for\,x = 0} \end{array}} \right.

Answer 1

We will find the left-hand limit and rind hand limit of the function at $x = 0$, using the L’Hospital rule for indeterminate form. Then we will compare it with the value of the function $x = 0$ to determine the type of discontinuity and if it will be a removable type of discontinuity we will rewrite to make it continuous.

Complete step-by-step answer:
Given data: f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{{4^x} - {e^x}}}{{{6^x} - 1}}\,for\,x \ne 0} \\\ {\log \left( {\dfrac{2}{3}} \right)\,for\,x = 0} \end{array}} \right.
We know that a function is continuous at a given point if and only if the left-hand limit, right-hand limit, and the value of the function at that point all are equal.
The left-hand limit of the function at $x = 0$
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 - h)$
Substituting the value of the function, we get
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{{4^x} - {e^x}}}{{{6^x} - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{4^{ - h}} - {e^{ - h}}}}{{{6^{ - h}} - 1}}$
Now here substituting $h = 0$ we are getting $\dfrac{0}{0}$ form
So we will use the L’hospital rule which states that if $\mathop {\lim }\limits_{x \to a} \dfrac{{h(x)}}{{g(x)}}$ is of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form then
$\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{h(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{h'(x)}}{{g'(x)}}$
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{4^{ - h}} - {e^{ - h}}}}{{{6^{ - h}} - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \left( {\log 4} \right){4^{ - h}} + {e^{ - h}}}}{{ - \left( {\log 6} \right){6^{ - h}}}}$
Now substituting $h = 0$, we get
$= \dfrac{{ - \left( {\log 4} \right){4^0} + {e^0}}}{{ - \left( {\log 6} \right){6^0}}}$
We know that ${a^0} = 1$, so using this, we get
$= \dfrac{{ - \left( {\log 4} \right) + 1}}{{ - \left( {\log 6} \right)}}$
Using $\log e = 1$ and $\log A - \log B = \log \left( {\dfrac{A}{B}} \right)$ , we get
$= \dfrac{{1 - \log 4}}{{ - \left( {\log 6} \right)}}$
On multiplying numerator and denominator with (-1), we get
$= \dfrac{{\log 4 - 1}}{{\log 6}}$

Now, the right-hand limit of the function at $x = 0$
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 + h)$
Substituting the value of the function, we get
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{4^x} - {e^x}}}{{{6^x} - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{4^h} - {e^h}}}{{{6^h} - 1}}$
Now here substituting $h = 0$ we are getting $\dfrac{0}{0}$ form
So we will use the L’hospital rule which states that if $\mathop {\lim }\limits_{x \to a} \dfrac{{h(x)}}{{g(x)}}$ is of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form then
$\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{h(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{h'(x)}}{{g'(x)}}$
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{4^h} - {e^h}}}{{{6^h} - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\log 4} \right){4^h} - {e^h}}}{{\left( {\log 6} \right){6^h}}}$
Now substituting $h = 0$, we get
$= \dfrac{{\left( {\log 4} \right){4^0} - {e^0}}}{{\left( {\log 6} \right){6^0}}}$
We know that ${a^0} = 1$, so using this, we get
$= \dfrac{{\log 4 - 1}}{{\log 6}}$
Using $\log e = 1$, we get
$= \dfrac{{1 - \log 4}}{{ - \left( {\log 6} \right)}}$
On multiplying numerator and denominator with (-1), we get
$= \dfrac{{\log 4 - 1}}{{\log 6}}$
Now it is given that that $f(0) = \log \left( {\dfrac{2}{3}} \right)$
Since the left-hand limit is equal to the right-hand limit, but they are not equal to the value of function removable discontinuity at $x = 0$
Redefining the function so that the function becomes continuous at $x = 0$
f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{{4^x} - {e^x}}}{{{6^x} - 1}}\,for\,x \ne 0} \\\ {\dfrac{{\log 4 - 1}}{{\log 6}}\,for\,x = 0} \end{array}} \right.

Note: While using the L’hospital rule i.e. if $\mathop {\lim }\limits_{x \to a} \dfrac{{h(x)}}{{g(x)}}$ is of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form then
$\mathop {\lim }\limits_{x \to a} \dfrac{{h(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{h'(x)}}{{g'(x)}}$ , most of the students differentiate the whole $\dfrac{{h(x)}}{{g(x)}}$ taking it as a single function which should not be done, we have to differentiate both numerator and the denominator separately, so remember this point to make a more accurate solution.