Question

Examine the continuity of the following function at given point
(i)f\left( x \right) = \left\\{ {\begin{array}{*{20}{c}} { = \dfrac{{{e^{5x}} - {e^{2x}}}}{{\sin 3x}}}&{{\text{for }}x \ne 0} \\\ { = 1}&{{\text{for }}x = 0} \end{array}} \right.

Answer 1

In this question, we have defined the given function for different values.
We need to find out the continuity of the function at the given point for finding that we first need to evaluate the left and right hand limit of the function at that point then if these two values are the same, we can conclude that the function is continuous at that point.

Formula used: A function $f$ is said to be continuous at a point $x = a$ if,
$\mathop {\lim }\limits_{x \to a - } f\left( x \right) = \mathop {\lim }\limits_{x \to a + } f\left( x \right) = f\left( a \right)$

Complete step-by-step solution:
It is given that the function $f$ is defined as,

{ = \dfrac{{{e^{5x}} - {e^{2x}}}}{{\sin 3x}}}&{{\text{for }}x \ne 0} \\\ { = 1}&{{\text{for }}x = 0} \end{array}} \right.$$ We need to examine the continuity of the following function at a given point $$x = 0$$. (The left hand limit of the function $f$ at $$x = 0$$) $$ = $$ (The right hand limit of the function $f$ at $$x = 0$$= The value of the function at $$x = 0$$) $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{2x}}}}{{\sin 3x}}$$ Substitute the limit, $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5 \times 0}} - {e^{2 \times 0}}}}{{\sin \left( {3 \times 0} \right)}}$$ Simplifying the terms, $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^0} - {e^0}}}{{\sin 0}}$$ This is in $$\dfrac{0}{0}$$ form thus we can apply L'Hospital's rule, we get, $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{2x}}}}{{\sin 3x}}$$ Simplifying we get, $$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{5{e^{5x}} - 2{e^{2x}}}}{{3\cos 3x}}$$ Hence, $$ \Rightarrow \dfrac{{5 - 2}}{3}$$ Subtracting the terms, $$ \Rightarrow \dfrac{3}{3}$$ Thus, $$ \Rightarrow 1$$ Then $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{2x}}}}{{\sin 3x}} = f\left( 0 \right) = 1$$ **$\therefore $ The function $f$ is continuous at the point $$x = 0$$.** **Note:** We have to know that, Continuous function: In mathematics, a continuous function is a function that does not have any abrupt changes in value, known as discontinuities. More precisely, sufficiently small changes in the input of a continuous function result in arbitrarily small changes in its output. If not continuous, a function is said to be discontinuous. L'Hospital's rule: It says that the limit when we divide one function by another is the same after we take the derivative of each function. $$\mathop {\lim }\limits_{x \to c} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$$ The limit as x approaches c of "$$f\left( x \right)$$over$$g\left( x \right)$$" equals the limit as x approaches c of "$$f'\left( x \right)$$ over$$g'\left( x \right)$$".