Question

For the given function $f\left( x \right) = {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}}$ if $x \ne a$ and $f\left( x \right) = k$ if $x = a$ and $f\left( x \right)$ is continuous at $x = a$ then $k =$
(A) ${e^{\tan a}}$
(B) ${e^{\cot a}}$
(C) ${e^a}$
(D) ${e^{1/a}}$

Answer 1

Hint : This question is based on Limits and Derivatives. The condition of the continuity of a function is that the limit of the function should exist. To solve this question, we have to arrange this expression in such a way that it satisfies the condition of the continuity and then substitute the limit value of the function and find the value of $k$ .

Complete step-by-step answer :
Given:
$f\left( x \right) = {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}}$ if $x \ne a$ and,
$f\left( x \right) = k$ if $x = a$
Also, it is given that $f\left( x \right)$ is continuous at $x = a$ . It means that-
$\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = k$
If we substitute $x = a$ in this equation we get,
$f\left( a \right) = {\left( {\dfrac{{\sin a}}{{\sin a}}} \right)^{\dfrac{1}{{a - a}}}}\\\ f\left( a \right) = {1^\infty }$
We cannot solve this expression by simple substitution. So, we have to write this expression in such a way that it can be solved easily. Now we know that if any expression reduces to ${1^\infty }$ form we can use this formula –
$\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$
So now, substituting the values into this formula we have –
$\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\mathop {\lim }\limits_{x \to a} \dfrac{1}{{x - a}}\left( {\dfrac{{\sin x}}{{\sin a}} - 1} \right)}}\\\ \mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\mathop {\lim }\limits_{x \to a} \dfrac{1}{{x - a}}\left( {\dfrac{{\sin x - \sin a}}{{\sin a}}} \right)}}$
We know the formula for $\left( {\sin x - \sin a} \right) = 2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)$
So, substituting this formula in the equation we get,
$\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{1}{{\sin a}}\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{x - a}}} \right)}}\\\ \mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{1}{{\sin a}}\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{\left( {\dfrac{{x - a}}{2}} \right)}}} \right)}}$
Now we know the rule that
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
So now we have to change the limit, then assume $x - a = t{\rm{ (suppose)}}$
The limit $x \to a$ changes to $x - a \to 0$ or $t \to 0$
Now we can apply this rule in the expression above. So, we have-
$\mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{{\cos \left( {\dfrac{{a + a}}{2}} \right)}}{{\sin a}}\left( {\mathop {\lim }\limits_{t \to 0} \dfrac{{\sin \left( {\dfrac{t}{2}} \right)}}{{\left( {\dfrac{t}{2}} \right)}}} \right)}}\\\ \mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\dfrac{{\cos a}}{{\sin a}}\left( {\mathop {\lim }\limits_{t \to 0} \dfrac{{\sin \left( {\dfrac{t}{2}} \right)}}{{\left( {\dfrac{t}{2}} \right)}}} \right)}}\\\ \mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\cot a \times 1}}\\\ \mathop {\lim }\limits_{x \to a} {\left( {\dfrac{{\sin x}}{{\sin a}}} \right)^{\dfrac{1}{{x - a}}}} = {e^{\cot a}}$
Therefore, the value of $k$ is ${e^{\cot a}}$
So, the correct answer is “Option B”.

Note : It should be noted that for a function to be continuous the limit of the function should exist and the value of the Right-hand limit should be equal to the Left-hand limit.
For example, if a function is continuous at point $x = a$ then
$\mathop {\lim }\limits_{x \to a} f\left( x \right)$ exists and
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$