Question

Evaluate: \[\mathop {\lim }\limits_{x \to 2} \dfrac{{{3^x} + {3^{3 - x}} - 12}}{{{3^{ - \dfrac{x}{2}}} - {3^{1 - x}}}}\]

Answer 1

Hint Simplify the given expression using algebraic identities and the law of exponents. Then substitute \[{3^{\dfrac{x}{2}}} = y\]. After that the given expression reduces to a function of \[y\]. \[x \to 2 \Rightarrow y \to 3\]. Put \[y = 3\] after canceling the term \[\left( {y - 3} \right)\].

Formula Used:
\[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\]
\[{a^{ - m}} = \dfrac{1}{{{a^m}}}\]
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step by step solution:
The given expression is \[\dfrac{{{3^x} + {3^{3 - x}} - 12}}{{{3^{ - \dfrac{x}{2}}} - {3^{1 - x}}}}\]
Use the identity \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\] and \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\]
Then the expression becomes
\[ = \dfrac{{{3^x} + \dfrac{{{3^3}}}{{{3^x}}} - 12}}{{\dfrac{1}{{{3^{\dfrac{x}{2}}}}} - \dfrac{{{3^1}}}{{{3^x}}}}}\]
\[ = \dfrac{{\left( {\dfrac{{{{\left( {{3^x}} \right)}^2} + 27 - 12 \cdot {3^x}}}{{{3^x}}}} \right)}}{{\left( {\dfrac{{{3^{\dfrac{x}{2}}} - 3}}{{{3^x}}}} \right)}}\]
\[ = \dfrac{{{3^{2x}} + 27 - 12 \cdot {3^x}}}{{{3^{\dfrac{x}{2}}} - 3}}\]
Let us substitute \[{3^{\dfrac{x}{2}}} = y\]
Then \[{3^{2x}} = {\left( {{3^{\dfrac{x}{2}}}} \right)^4} = {y^4}\] and \[{3^x} = {\left( {{3^{\dfrac{x}{2}}}} \right)^2} = {y^2}\]
Substitute these in the above expression.
\[ = \dfrac{{{y^4} + 27 - 12{y^2}}}{{y - 3}}\]
\[ = \dfrac{{{y^4} - 12{y^2} + 27}}{{y - 3}}\]
The numerator is a quadratic polynomial in \[{y^2}\].
Factorize the polynomial.
\[{y^4} - 12{y^2} + 27\]
\[ = {y^4} - 9{y^2} - 3{y^2} + 27\]
\[ = {y^2}\left( {{y^2} - 9} \right) - 3\left( {{y^2} - 9} \right)\]
\[ = \left( {{y^2} - 9} \right)\left( {{y^2} - 3} \right)\]
Use the identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
\[ = \left( {y + 3} \right)\left( {y - 3} \right)\left( {{y^2} - 3} \right)\]
So, the expression becomes
\[ = \dfrac{{\left( {y + 3} \right)\left( {y - 3} \right)\left( {{y^2} - 3} \right)}}{{\left( {y - 3} \right)}}\]
Given that \[x \to 2\]
At \[x = 2\], \[y = 3\]
So, \[x \to 2 \Rightarrow y \to 3\]
and \[y \to 3 \Rightarrow y \ne 3\]
So, \[y - 3 \ne 0\]
Cancel the term \[\left( {y - 3} \right)\] from numerator and denominator.
\[\therefore \]\[\mathop {\lim }\limits_{x \to 2} \dfrac{{{3^x} + {3^{3 - x}} - 12}}{{{3^{ - \dfrac{x}{2}}} - {3^{1 - x}}}} = \mathop {\lim }\limits_{y \to 3} \left( {y + 3} \right)\left( {{y^2} - 3} \right) = \left( {3 + 3} \right)\left( {{3^2} - 3} \right) = 6 \times 6 = 36\]
Hence, the required value is \[36\]

Note: Many students get confused seeing such a function. They should not be confused about it. Just simplify the expression using some laws, some identities. After simplification you’ll get a simple function. Then find out the required solution by taking the simple function.