$$\lim_{x \to 2} \frac{3^x + 3^{3-x} - 12}{3^{-x/2} - 3^{1-x}}$$ is equal to

We have to evaluate the limit

$$$\displaystyle \lim_{x \to 2}\frac{3^{x}+3^{\,3-x}-12}{3^{-x/2}-3^{\,1-x}}.$$$

First we check the value of the numerator and the denominator at $$x=2$$ to identify the indeterminate form.

For the numerator,

$$$3^{x}+3^{\,3-x}-12 \; \Big|_{x=2}=3^{2}+3^{\,3-2}-12=9+3-12=0.$$$

For the denominator,

$$$3^{-x/2}-3^{\,1-x}\; \Big|_{x=2}=3^{-2/2}-3^{\,1-2}=3^{-1}-3^{-1}=0.$$$

So the limit is of the indeterminate type $$0/0$$. We therefore need to simplify the expression. A convenient way is to shift the variable so that the point $$x=2$$ becomes $$y=0$$.

Put $$x=2+y,$$ where $$y \to 0$$ as $$x \to 2$$. We now express every power of 3 in terms of $$y$$.

Using the law $$3^{a+b}=3^{a}\,3^{b},$$ the numerator becomes

$$$\begin{aligned} 3^{x}+3^{\,3-x}-12 &=3^{\,2+y}+3^{\,3-(2+y)}-12 \\ &=3^{2}\,3^{y}+3^{1}\,3^{-y}-12 \\ &=9\,3^{y}+3\,3^{-y}-12. \end{aligned}$$$

The denominator is

$$$\begin{aligned} 3^{-x/2}-3^{\,1-x} &=3^{-(2+y)/2}-3^{\,1-(2+y)} \\ &=3^{-1-y/2}-3^{-1-y} \\ &=3^{-1}\,3^{-y/2}-3^{-1}\,3^{-y} \\ &=\frac13\bigl(3^{-y/2}-3^{-y}\bigr). \end{aligned}$$$

Dividing by $$\frac13$$ is equivalent to multiplying the whole fraction by 3, so the complete expression becomes

$$$\begin{aligned} \frac{3^{x}+3^{\,3-x}-12}{3^{-x/2}-3^{\,1-x}} &=\frac{9\,3^{y}+3\,3^{-y}-12}{\frac13\bigl(3^{-y/2}-3^{-y}\bigr)} \\ &=3\,\frac{9\,3^{y}+3\,3^{-y}-12}{3^{-y/2}-3^{-y}}. \end{aligned}$$$

Now we use the exponential expansion around $$y=0$$: for a small $$y$$,

$$3^{y}=1+y\ln3+\frac{y^{2}(\ln3)^{2}}{2}+\cdots,$$

$$3^{-y}=1-y\ln3+\frac{y^{2}(\ln3)^{2}}{2}+\cdots,$$

$$3^{-y/2}=1-\frac{y}{2}\ln3+\frac{y^{2}(\ln3)^{2}}{8}+\cdots.$$

Substituting these in the numerator:

$$$\begin{aligned} 9\,3^{y}+3\,3^{-y}-12 &=9\!\left(1+y\ln3+\frac{y^{2}(\ln3)^{2}}{2}\right) +3\!\left(1-y\ln3+\frac{y^{2}(\ln3)^{2}}{2}\right) -12 \\ &=(9+3-12) \;+\; (9-3)\,y\ln3 \;+\; \left(\frac{9}{2}+\frac{3}{2}\right)\!y^{2}(\ln3)^{2}+\cdots \\ &=0+6y\ln3+6y^{2}(\ln3)^{2}+\cdots. \end{aligned}$$$

So, to first order,

$$\text{Numerator}=6y\ln3+O(y^{2}).$$

For the denominator, we expand similarly:

$$$\begin{aligned} 3^{-y/2}-3^{-y} &=\left(1-\frac{y}{2}\ln3+\frac{y^{2}(\ln3)^{2}}{8}\right) -\left(1-y\ln3+\frac{y^{2}(\ln3)^{2}}{2}\right) \\ &=\left(-\frac{y}{2}\ln3\right)+y\ln3 +\left(\frac{1}{8}-\frac12\right)y^{2}(\ln3)^{2}+\cdots \\ &=\frac{y\ln3}{2}-\frac{3y^{2}(\ln3)^{2}}{8}+\cdots. \end{aligned}$$$

Thus, to first order,

$$\text{Denominator}=\frac{y\ln3}{2}+O(y^{2}).$$

Forming their ratio we get

$$$\frac{9\,3^{y}+3\,3^{-y}-12}{3^{-y/2}-3^{-y}} =\frac{6y\ln3+O(y^{2})}{\dfrac{y\ln3}{2}+O(y^{2})} =\frac{6+O(y)}{\dfrac12+O(y)} \longrightarrow\frac{6}{1/2}=12 \quad\text{as }y\to0.$$$

Remember we still have the factor 3 outside, hence

$$$\lim_{x \to 2}\frac{3^{x}+3^{\,3-x}-12}{3^{-x/2}-3^{\,1-x}} =3\times 12=36.$$$

So, the answer is $$36$$.