Question

$limi{t}_{x\to 2}\left(\frac{3^x+3^{3-x}-12}{3^{-{\displaystyle \frac{x}{2}}}-3^{1-x}}\right)=$

Answer 1

Step 1: Substituting $y$

Let we choose $y=3^{\frac{x}{2}}\Rightarrow y^2=3^{\frac{2x}{2}}=3^x$

Limit $x\to 2$, $\Rightarrow y\to 3\left[\because y=3^{\frac{2}{2}}=3\right]$

Step 3:Evaluating the given function:

Therefore,

$$\begin{gathered} limi{t}_{x\to 2}\left(\frac{3^x+{\displaystyle \frac{3^3}{3^x}}-12}{{\displaystyle \frac{1}{3^{{\displaystyle \frac{x}{2}}}}-\frac{3}{3^x}}}\right)=limi{t}_{y\to 3}\left(\frac{y^2+{\displaystyle \frac{3^3}{y^2}}-12}{{\displaystyle \frac{1}{y}-\frac{3}{y^2}}}\right)\mathbf{}\left[\because y=3^{\frac{x}{2}}\Rightarrow y^2=3^{\frac{2x}{2}}=3^x\right] \\ =limi{t}_{y\to 3}\left(\frac{\cancel{{\displaystyle \frac{1}{y^2}}}\left(y^4+{\displaystyle 27}-12y^2\right)}{{\displaystyle \cancel{\frac{1}{y^2}}\left(y-3\right)}}\right)\mathbf{}\mathbf{[}\mathbf{}\mathbf{\text{taking}}\mathbf{}\frac{\mathbf{1}}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{\text{common in numerator and denominator}}\mathbf{]} \\ =limi{t}_{y\to 3}\frac{\left(y^4+{\displaystyle 27}-12y^2\right)}{{\displaystyle \left(y-3\right)}} \end{gathered}$$Factorize the numerator

$$\begin{gathered} =limi{t}_{y\to 3}\frac{\left(y^4-3y^2-9y^2+27\right)}{{\displaystyle \left(y-3\right)}} \\ =limi{t}_{y\to 3}\frac{y^2\left(y^2-3\right)-9\left(y^2-3\right)}{{\displaystyle \left(y-3\right)}} \\ =limi{t}_{y\to 3}\frac{\left(y^2-9\right)\left(y^2-3\right)}{{\displaystyle \left(y-3\right)}} \\ =limi{t}_{y\to 3}\frac{\left(y^2-3\right)\cancel{\left(y-3\right)}\left(y+3\right)}{{\displaystyle \cancel{\left(y-3\right)}}} \\ =\left(3+3\right)\left(3^2-3\right) \\ =36 \end{gathered}$$

Hence,${\mathbf{limit}}_{\mathbf{x}\mathbf{\to }\mathbf{2}}\left(\frac{3^x+3^{3-x}-12}{3^{-{\displaystyle \frac{x}{2}}}-3^{1-x}}\right)\mathbf{=}\mathbf{36}$