Question

If we are given the function as \[f(x)=\left\{ \begin{align}
  & \dfrac{{1 - \cos 4x}}{{{x^2}}},{\text{ when }}x < 0 \\
 & {\text{ a, when }}x = 0 \\
& \dfrac{{\sqrt x }}{{\sqrt {(16 + \sqrt x )} - 4}},{\text{ when }}x > 0
\end{align} \right.\]
 is continuous at \[x = 0\] ,then the value of \[a\] will be?
(A) \[8\]
(B) \[ - 8\]
(C) \[4\]
(D) \[{\text{None of these}}\]

Answer 1

Hint: To solve this problem, we use concepts based on limits and continuity. We will use some definitions of continuity to solve this problem. We may use the formula, \[\mathop {\lim }\limits_{x \to a} f(x) = f(a)\] to evaluate our problem. Also, we use some trigonometric identities and properties of limits to solve this problem.

Complete step-by-step solution:
Limit of a function is defined as a value, at which the function is converging for the given point.
Written as \[\mathop {\lim }\limits_{x \to a} f(x)\] which means, limit of function \[f(x)\] as \[x\] approaches \[a\] .
Let us consider a function \[f(x)\] which is defined in the range \[\left[ {a,b} \right]\] . So, this function is continuous at \[x = c\] where \[c \in \left[ {a,b} \right]\] , if and only if, both the left limit and the right limit exist and are equal.
That means if \[\mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = l\] where \[l\] is a finite value.
And also \[l\] can be defined as \[f(c)\] .
So, in the problem, \[f(x)\] is continuous at \[x = 0\] that means, \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)\] .
So, now let’s consider only the left limit for our convenience.
So, \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 4x}}{{{x^2}}} = f(0)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - (1 - 2{{\sin }^2}2x)}}{{{x^2}}} = f(0)\] -----from the identity, \[\cos 2x = 1 - 2{\sin ^2}x\] .
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}2x}}{{{x^2}}} = a\] -----we know the value \[f(0) = a\] .
Now, multiply both the numerator and denominator by \[4\] .
So, we get, \[\mathop {\lim }\limits_{x \to 0} \dfrac{{2(4){{\sin }^2}2x}}{{(4){x^2}}} = a\]
\[ \Rightarrow 8\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{(4{x^2})}} = a\]
\[ \Rightarrow 8\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}(2x)}}{{{{(2x)}^2}}} = a\]
Now, as both numerator and denominator are in squares, we can take them as one single square.
\[ \Rightarrow 8\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin 2x}}{{2x}}} \right)^2} = a\]
Limits can be applied or shifted into the bracket. So, we get,
\[ \Rightarrow 8{\left( {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{2x}}} \right)^2} = a\]
In limits, we have an identity which is as follows \[\mathop {\lim }\limits_{n \to 0} \dfrac{{\sin n}}{n} = 1\] .
Here, \[{\text{as }}x \to 0 \Rightarrow 2x \to 0\] ---by multiplying both by two.
So, from this formula, we get, \[8{\left( 1 \right)^2} = a\]
\[ \Rightarrow a = 8\]
So, option (A) is the correct option.

Note: Here, we have considered only the left limit to solve the problem. But we can also consider the right limit too. And the solution will be as follows.
\[\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt x }}{{\sqrt {(16 + \sqrt x )} - 4}} = f(0)\]
We use L'Hospital's rule to solve this sum, which means, if we are getting a fraction of kind \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] after substituting the point in the limit, we differentiate the numerator and denominator separately, and then substitute the point value in the limit.
Here, we are getting \[\dfrac{0}{0}\] after substituting \[x = 0\] , so we differentiate numerator and denominator separately.
So, we get,
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{2\sqrt x }}}}{{\dfrac{1}{{2\sqrt {(16 + \sqrt x )} }} \times \dfrac{1}{{2\sqrt x }}}} = a\] -----\[\left( {\because \dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} 2\sqrt {(16 + \sqrt x )} = a\]
Now we can substitute \[x = 0\] and we get, \[a = 8\]
So, either way we can solve this question.