Question

Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous
\[f(x)=\left\{ \begin{align} & \dfrac{{e^{5\sin x}}-e^{2x}}{5\tan(x)-3x}, for x\ne 0 \\ & \dfrac{3}{4}, x=0 \\ \end{align} \right.\]

Answer 1

Hint: Removable discontinuity is defined as one at which the limit of the function exists but does not equal to the value of the function at that point and this may be because the function does not exist at that point. Firstly, we will check that the function has removable continuity and if it has a removable discontinuity, we defined it so that it becomes continuous by using the limits formulas.

Complete step by step answer:
In the given question firstly, we have to determine that the given function has removable discontinuity and then we will redefine the function so that it becomes a continuous function.
Here, the given function has removable discontinuity as $\mathop {\lim }\limits_{x \to 0} f(x) \ne f(0)$
So, now will redefine the function.
We have $f(0) = \dfrac{3}{4}$
Now, \[\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5\sin x}} - {e^{2x}}}}{{5\tan x - 3x}}\]
We can rewrite the above equation as
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{({e^{5\sin x}} - 1) - ({e^{2x}} - 1)}}{{5\tan x - 3x}}\]
Divide numerator and denominator by $x$ as $x \to 0$ and also $x \ne 0$. So,
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\dfrac{{({e^{5\sin x}} - 1) - ({e^{2x}} - 1)}}{x}}}{{\dfrac{{5\tan x - 3x}}{x}}}} \right]\]
We can rewrite the above equation as
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = \dfrac{{\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{({e^{5\sin x}} - 1)}}{x}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{({e^{2x}} - 1)}}{x}} \right]}}{{\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{5\tan x}}{x}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{3x}}{x}} \right]}}\]
On multiplying and dividing the term $\dfrac{{{e^{5\sin x}} - 1}}{x}$ by $5\sin x$ and the term $\dfrac{{{e^{2x}} - 1}}{x}$ by $2$. We get,
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = \dfrac{{\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{({e^{5\sin x}} - 1)}}{{5\sin x}}\dfrac{{5\sin x}}{x}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{({e^{2x}} - 1)}}{{2x}} \times 2} \right]}}{{\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{5\tan x}}{x}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ 3 \right]}}\]
The above equation can be written as
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = \dfrac{{5\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{5\sin x}} - 1}}{{5\sin x}}} \right) \cdot \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) - 2\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{({e^{2x}} - 1)}}{{2x}}} \right)}}{{5\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\tan x}}{x}} \right) - \mathop {\lim }\limits_{x \to 0} (3)}}\]
As $x \to 0,\,\,2x \to 0,\,\,\sin x \to 0,\,\,5\sin x \to 0$. So, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^x} - 1}}{x}} \right) = 1,\,\,\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
Therefore,
$ \Rightarrow \dfrac{{5(1)(1) - 2(1)}}{{5(1) - 3}}$
On multiplying we get,
$ \Rightarrow \dfrac{{5 - 2}}{{5 - 3}}$
On subtracting we get,
$ \Rightarrow \dfrac{3}{2}$
$\therefore \mathop {\lim }\limits_{x \to 0} f(x) \ne f(0)$
$\therefore f(x)$ is continuous at $x = 0$
$\therefore f(x)$ has a removable discontinuity at $x = 0$
Therefore, the discontinuity can be removed by redefining the function $f(0) = \dfrac{3}{2}$
Hence, the function can be defined as
\[f(x)=\left\{ \begin{align} & \dfrac{{e^{5\sin x}}-e^{2x}}{5\tan(x)-3x}, for x\ne 0 \\ & \dfrac{3}{2}, x=0 \\ \end{align} \right.\]

Note:
Note that a discontinuity can be removable at a point $x = a$ if the $\mathop {\lim }\limits_{a \to 0} f(x)$ exists and this limit is finite and there are two types of removable discontinuities which are the function is undefined at $x = a$ and the value of the function at $x = a$ does not match the limit and when a function has a removable discontinuity it can be defined to make it continuous function.