Question

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

Answer 1

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,

& \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{aligned}$$ Applying the formula: - $$\log {{a}^{m}}=m\log a$$, we get, $$\begin{aligned} & \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{aligned}$$ Applying the formula: - $${{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{aligned} & \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{aligned}$$ **So, the correct answer is “Option B”.** **Note:** One may note that we can also evaluate the limit without using L – Hospital Rule. We can divide the numerator and denominator with 16 and convert the function in the form $$\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 }$$.