If the function, \[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16},for x\ne 2 \\
   A,for x = 2 \\
\end{matrix} \right.\], is continuous at x = 2, then A =
(a) 2
(b) \[\dfrac{1}{2}\]
(c) \[\dfrac{1}{4}\]
(d) 0

Hint: Apply L – Hospital’s Rule to find the value of left hand limit = right hand limit = \[\underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}\] and substitute this value equal to \[f\left( 2 \right)=A\]. Use the theorem that, “if at x = a \[\dfrac{f\left( x \right)}{g\left( x \right)}\] is of \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] form then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}\]”.

Complete step by step answer:
We have been provided that, \[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16},forx\ne 2 \\
   A,forx\ne 2 \\
\end{matrix} \right.\], is continuous at x = 2 and we have to find the value of A.
Let us see the conditions if a function is continuous at some point x = a.
Now, if a function is continuous at x = a, then the values of left hand side limit (L.H.L), right hand limit (R.H.L) and f (a) must be the same. Mathematically,
\[\Rightarrow \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\]
In the above question, we have been given that: - x = 2, f (x) = A and at \[x\ne 2\], \[f\left( x \right)=\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}\]. So, mathematically,
\[\Rightarrow f\left( 2 \right)=A\] - (i)
\[\Rightarrow \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,=\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}\] - (ii)
So, let us find the value of relation (ii),
\[\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}\], here when we will substitute x = 2 in \[\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}\] then it will convert into \[\dfrac{0}{0}\] form. So, L – Hospital’s Rule can be applied here.
L – Hospital’s Rule states that, “if a function is of the form \[\dfrac{f\left( x \right)}{g\left( x \right)}\] and at x = a it is of the form \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}\]”, where \[f'\left( x \right)\] and \[g'\left( x \right)\] are the derivatives of \[f\left( x \right)\] and \[g\left( x \right)\] respectively.
So, considering \[{{2}^{x+2}}-16=f\left( x \right)\] and \[{{4}^{x}}-16=g\left( x \right)\], we get,
\[\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\underset{x\to 2}{\mathop{\lim }}\,\dfrac{\dfrac{d\left[ {{2}^{x+2}}-16 \right]}{dx}}{\dfrac{d\left[ {{4}^{x}}-16 \right]}{dx}}\]
We know that, \[\dfrac{d\left[ {{a}^{x}} \right]}{dx}={{a}^{x}}\log a\] and derivative of constant is 0. Therefore, we have,
\[\begin{align}
  & \Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\dfrac{{{2}^{x+2}}\log 2-0}{{{4}^{x}}\log 4-0} \\
 & \Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\dfrac{{{2}^{x+2}}\log 2}{{{4}^{x}}\log 4} \\
 & \Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\dfrac{{{2}^{x+2}}\log 2}{{{2}^{2x}}\left( \log {{2}^{2}} \right)} \\
\end{align}\]
Applying the formula: - \[\log {{a}^{m}}=m\log a\], we get,
\[\begin{align}
  & \Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\dfrac{{{2}^{x+2}}\log 2}{{{2}^{2x}}\times 2\log 2} \\
 & \Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\dfrac{{{2}^{x+2}}}{{{2}^{2x}}\times 2} \\
\end{align}\]
Applying the formula: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] and \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], we get,
\[\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=\underset{x\to 2}{\mathop{\lim }}\,{{2}^{\left( x+2 \right)-\left( 1+2x \right)}}=\underset{x\to 2}{\mathop{\lim }}\,{{2}^{1-x}}={{2}^{1-2}}={{2}^{-1}}=\dfrac{1}{2}\]
Since, the function is continuous, therefore,
\[\begin{align}
  & \Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}=f\left( 2 \right) \\
 & \Rightarrow {{2}^{-1}}=A \\
 & \Rightarrow A=\dfrac{1}{{{2}^{1}}} \\
 & \Rightarrow A=\dfrac{1}{2} \\
\end{align}\]

So, the correct answer is “Option B”.

Note: One may note that we can also evaluate the limit without using L – Hospital Rule. We can divide the numerator and denominator with 16 and convert the function in the form \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}\] whose solution is \[{{\log }_{e}}a\]. But we have applied L – Hospital’s Rule because it is easier to solve the limit problem with this theorem. The most important thing is that L – Hospital’s Rule is applicable for limits of the form \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\].