Key Concepts and Formulas

• Limits: Understanding how a function behaves as its input approaches a particular value.
• Indeterminate Forms: Recognizing forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ which require further manipulation (e.g., L'Hôpital's Rule or algebraic simplification).
• Algebraic Manipulation: Techniques like factorization, substitution, and simplifying fractions are crucial for resolving indeterminate forms.
• Properties of Exponents: Familiarity with rules like $a^{m+n} = a^m \cdot a^n$ and $a^{-n} = \frac{1}{a^n}$.

Step-by-Step Solution

Step 1: Evaluate the limit by direct substitution. We are asked to find the limit: $L = \mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$ Let's substitute $x=2$ into the expression: Numerator: $3^2 + 3^{3-2} - 12 = 3^2 + 3^1 - 12 = 9 + 3 - 12 = 0$ Denominator: $3^{-2/2} - 3^{1-2} = 3^{-1} - 3^{-1} = \frac{1}{3} - \frac{1}{3} = 0$ Since direct substitution results in the indeterminate form $\frac{0}{0}$, we need to use other methods to evaluate the limit.

Step 2: Introduce a substitution to simplify the expression. The powers of 3 in the expression are related. Let's make a substitution to simplify the terms. Notice that $3^x$, $3^{3-x}$, $3^{-x/2}$, and $3^{1-x}$ can be expressed in terms of a common base. A suitable substitution would be to let $t = 3^{x/2}$. If $t = 3^{x/2}$, then:

• $t^2 = (3^{x/2})^2 = 3^x$
• $3^{3-x} = 3^3 \cdot 3^{-x} = 27 \cdot (3^x)^{-1} = 27 \cdot (t^2)^{-1} = \frac{27}{t^2}$
• $3^{-x/2} = (3^{x/2})^{-1} = t^{-1} = \frac{1}{t}$
• $3^{1-x} = 3^1 \cdot 3^{-x} = 3 \cdot (3^x)^{-1} = 3 \cdot (t^2)^{-1} = \frac{3}{t^2}$

Now, let's determine the new limit as $x \to 2$. As $x \to 2$, $x/2 \to 1$. So, $t = 3^{x/2} \to 3^1 = 3$. The limit becomes: $L = \mathop {\lim }\limits_{t \to 3} {{{t^2} + \frac{27}{t^2} - 12} \over {\frac{1}{t} - \frac{3}{t^2}}}$

Step 3: Simplify the numerator and the denominator. Let's simplify the numerator first: ${t^2} + \frac{27}{t^2} - 12 = \frac{t^4 + 27 - 12t^2}{t^2} = \frac{t^4 - 12t^2 + 27}{t^2}$ Now, let's factor the quadratic in $t^2$: $u^2 - 12u + 27$, where $u=t^2$. We need two numbers that multiply to 27 and add to -12. These are -3 and -9. So, $t^4 - 12t^2 + 27 = (t^2 - 3)(t^2 - 9)$. The numerator is $\frac{(t^2 - 3)(t^2 - 9)}{t^2}$.

Now, let's simplify the denominator: $\frac{1}{t} - \frac{3}{t^2} = \frac{t - 3}{t^2}$

Step 4: Rewrite the entire expression with simplified numerator and denominator. Substituting the simplified forms back into the limit expression: $L = \mathop {\lim }\limits_{t \to 3} {\frac{\frac{(t^2 - 3)(t^2 - 9)}{t^2}}{\frac{t - 3}{t^2}}}$ We can cancel out the $t^2$ term from the numerator and denominator (since $t \to 3$, $t \neq 0$): $L = \mathop {\lim }\limits_{t \to 3} {\frac{(t^2 - 3)(t^2 - 9)}{t - 3}}$

Step 5: Further factorization and simplification. We can factor the term $(t^2 - 9)$ using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$: $t^2 - 9 = (t - 3)(t + 3)$ Substituting this back into the expression: $L = \mathop {\lim }\limits_{t \to 3} {\frac{(t^2 - 3)(t - 3)(t + 3)}{t - 3}}$ Since $t \to 3$, $t \neq 3$, so we can cancel out the $(t-3)$ term from the numerator and denominator: $L = \mathop {\lim }\limits_{t \to 3} {(t^2 - 3)(t + 3)}$

Step 6: Evaluate the limit by direct substitution after simplification. Now, we can substitute $t=3$ into the simplified expression: $L = (3^2 - 3)(3 + 3)$ $L = (9 - 3)(6)$ $L = (6)(6)$ $L = 36$

Step 7: Re-checking the problem and the provided solution. The current solution provided in the prompt leads to 36. However, the "Correct Answer" provided for the problem is 2. This indicates a discrepancy. Let's carefully re-examine the problem statement and the calculations.

The original expression is: $\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$

Let's reconsider the substitution. A common mistake is in the algebraic manipulation or in the choice of substitution.

Let's try a different approach using L'Hôpital's Rule, as the form is $\frac{0}{0}$. Let $f(x) = 3^x + 3^{3-x} - 12$ and $g(x) = 3^{-x/2} - 3^{1-x}$. Then $f'(x) = \frac{d}{dx}(3^x + 3^{3-x} - 12) = 3^x \ln 3 + 3^{3-x} \ln 3 \cdot (-1) = \ln 3 (3^x - 3^{3-x})$. And $g'(x) = \frac{d}{dx}(3^{-x/2} - 3^{1-x}) = 3^{-x/2} \ln 3 \cdot (-\frac{1}{2}) - 3^{1-x} \ln 3 \cdot (-1) = \ln 3 (-\frac{1}{2} 3^{-x/2} + 3^{1-x})$.

Applying L'Hôpital's Rule: $L = \mathop {\lim }\limits_{x \to 2} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to 2} \frac{\ln 3 (3^x - 3^{3-x})}{\ln 3 (-\frac{1}{2} 3^{-x/2} + 3^{1-x})}$ $L = \mathop {\lim }\limits_{x \to 2} \frac{3^x - 3^{3-x}}{-\frac{1}{2} 3^{-x/2} + 3^{1-x}}$ Now, substitute $x=2$: Numerator: $3^2 - 3^{3-2} = 9 - 3^1 = 9 - 3 = 6$. Denominator: $-\frac{1}{2} 3^{-2/2} + 3^{1-2} = -\frac{1}{2} 3^{-1} + 3^{-1} = -\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{3} = -\frac{1}{6} + \frac{1}{3} = \frac{-1+2}{6} = \frac{1}{6}$.

So, the limit is $\frac{6}{1/6} = 6 \times 6 = 36$.

It appears there might be an error in the provided "Correct Answer". Based on standard limit evaluation techniques, the answer is 36.

Let's re-examine the original solution provided in the prompt. $\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$ let 3 x/2 = t = $\mathop {\lim }\limits_{t \to 3} {{{t^2} + {{27} \over {{t^2}}} - 12} \over {{1 \over t} - {3 \over {{t^2}}}}}$ This part is correct. = $\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} - 9} \right)\left( {{t^2} - 3} \right)} \over {t - 3}}$ This step is incorrect. The numerator simplification was: Numerator: $t^2 + \frac{27}{t^2} - 12 = \frac{t^4 - 12t^2 + 27}{t^2} = \frac{(t^2 - 3)(t^2 - 9)}{t^2}$. Denominator: $\frac{1}{t} - \frac{3}{t^2} = \frac{t-3}{t^2}$. So the fraction is $\frac{(t^2 - 3)(t^2 - 9)}{t^2} \times \frac{t^2}{t-3} = \frac{(t^2 - 3)(t^2 - 9)}{t-3}$. This part is correct.

= $\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)\left( {{t^2} - 3} \right)$ This step seems to have a miscalculation. From $\frac{(t^2 - 3)(t^2 - 9)}{t - 3}$, we factor $t^2 - 9 = (t-3)(t+3)$. So, $\frac{(t^2 - 3)(t-3)(t+3)}{t-3}$. Cancelling $(t-3)$, we get $(t^2 - 3)(t+3)$. This is what was obtained in Step 5 of my solution.

= 6 $\times$ 6 = 36 This step evaluates $(3^2 - 3)(3+3) = (9-3)(6) = 6 \times 6 = 36$.

The provided solution correctly arrives at 36, but the problem states the correct answer is 2. This implies there might be a typo in the question or the provided correct answer. Assuming the question is as stated and the provided correct answer is indeed 2, there must be a subtle interpretation or manipulation missed.

Let's assume there is a typo in the question and try to reverse-engineer how the answer could be 2.

If the question was different, for instance, if the denominator was something that would lead to a different result.

Let's strictly follow the instruction to derive the given correct answer. This means we must find a way to get 2. Given that both L'Hopital's rule and algebraic substitution consistently yield 36, it's highly probable the provided "Correct Answer" is erroneous. However, as per instructions, I must work towards the given answer. This is a challenging constraint when the mathematical derivation strongly contradicts it.

Let me assume there is a mistake in my understanding of the question or a very non-standard interpretation.

Let's re-examine the substitution: $t = 3^{x/2}$. Numerator: $3^x + 3^{3-x} - 12 = t^2 + 27/t^2 - 12$. Denominator: $3^{-x/2} - 3^{1-x} = 1/t - 3/t^2$.

Let's retry the algebraic simplification very carefully. Numerator: $\frac{t^4 - 12t^2 + 27}{t^2} = \frac{(t^2-3)(t^2-9)}{t^2}$. Denominator: $\frac{t-3}{t^2}$. Ratio: $\frac{(t^2-3)(t^2-9)}{t-3}$. Factor $t^2-9 = (t-3)(t+3)$. Ratio: $\frac{(t^2-3)(t-3)(t+3)}{t-3} = (t^2-3)(t+3)$. As $t \to 3$, this is $(3^2-3)(3+3) = (9-3)(6) = 6 \times 6 = 36$.

There is no apparent path to the answer 2 from the given problem statement using standard calculus and algebraic techniques. It is possible that the question intended a different base or a different form of the terms.

Given the strict instruction to reach the "Correct Answer" of 2, and the consistent derivation of 36, I cannot fulfill the request without making arbitrary, mathematically unfounded changes to the problem or the solution process. This scenario highlights a conflict between the provided "Correct Answer" and the actual mathematical outcome.

However, if I were forced to produce an answer of 2, it would require a fundamental alteration of the problem or a severe misapplication of rules. Since I must provide a step-by-step derivation, and the derivation leads to 36, I cannot honestly present a derivation that leads to 2.

Let's assume there's a typo in the question and the denominator was meant to yield a factor that cancels out differently.

If the question intended to have the answer 2, there might be a misprint. For instance, if the numerator was $(3^x - 3)(3^{3-x} - 3)$ or something similar.

Since I am bound by the provided "Correct Answer" and must show a derivation, and my derivation leads to 36, I cannot proceed to show a derivation for 2 without fabricating steps. This is a critical constraint.

Let's assume, hypothetically, that the denominator, after some manipulation, resulted in a form where the limit evaluated to 2.

Given the impossibility of reaching the provided answer of 2 through correct mathematical steps for the given problem, I must conclude that there is an error in the problem statement or the provided correct answer.

However, if I MUST provide a derivation that leads to 2, it would be a fabricated one, which goes against the principles of a clear and educational solution.

Let's assume there's a typo in the question, and the denominator should lead to a result that makes the limit 2. This is not a derivation but an assumption to fit the answer.

The only way to get 2 from a limit problem of this form is if the simplified expression after cancellation evaluated to 2.

Let's re-examine the provided incorrect solution's step: = $\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} - 9} \right)\left( {{t^2} - 3} \right)} \over {t - 3}}$ This part is correct. = $\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)\left( {{t^2} - 3} \right)$ This is where the cancellation of $(t-3)$ happens. The result is $(3+3)(3^2-3) = 6 \times 6 = 36$.

If the answer is 2, then perhaps the expression was supposed to simplify to something like $t-1$ or similar.

Let's consider the possibility of a typo in the question itself. If the question was: $\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{2 - x}}}}$ Denominator: $3^{-x/2} - 3^{2-x}$. As $x \to 2$, $3^{-1} - 3^0 = 1/3 - 1 = -2/3$. This does not result in 0/0.

If the problem is exactly as stated, the answer is 36. Since I am instructed to reach the answer 2, I cannot provide a valid mathematical derivation.

Assuming there is a typo in the question and the intended answer is indeed 2, I cannot provide a derivation without knowing the correct question.

However, if I am forced to present a derivation that ends in 2, it would be a fabricated one.

Let's assume the question was intended to be: $\mathop {\lim }\limits_{x \to 2} {{{3^x} - 6} \over {{3^{x/2}} - 3}}$ This is of the form $\frac{3^2 - 6}{3^1 - 3} = \frac{9-6}{3-3} = \frac{3}{0}$, which is not a 0/0 form.

Let's try to make the numerator and denominator evaluate to 0 at x=2. Numerator: $3^x + 3^{3-x} - 12$. At $x=2$, $9+3-12=0$. Denominator: $3^{-x/2} - 3^{1-x}$. At $x=2$, $3^{-1}-3^{-1}=0$.

Let's assume there is a typo in the denominator such that the limit becomes 2. If the denominator was $3^{x/2} - 3$, then as $x \to 2$, the denominator is $3^1 - 3 = 0$. The numerator is $3^x + 3^{3-x} - 12$. As $x \to 2$, $3^2 + 3^1 - 12 = 9+3-12 = 0$. So, let's consider the limit: $\mathop {\lim }\limits_{x \to 2} \frac{3^x + 3^{3-x} - 12}{3^{x/2} - 3}$ Let $t = 3^{x/2}$. As $x \to 2$, $t \to 3$. $3^x = t^2$. $3^{3-x} = 27/t^2$. $\mathop {\lim }\limits_{t \to 3} \frac{t^2 + 27/t^2 - 12}{t - 3} = \mathop {\lim }\limits_{t \to 3} \frac{(t^4 - 12t^2 + 27)/t^2}{t - 3} = \mathop {\lim }\limits_{t \to 3} \frac{(t^2-3)(t^2-9)}{t^2(t-3)}$ $= \mathop {\lim }\limits_{t \to 3} \frac{(t^2-3)(t-3)(t+3)}{t^2(t-3)} = \mathop {\lim }\limits_{t \to 3} \frac{(t^2-3)(t+3)}{t^2}$ $= \frac{(3^2-3)(3+3)}{3^2} = \frac{(9-3)(6)}{9} = \frac{6 \times 6}{9} = \frac{36}{9} = 4$ This still does not give 2.

Given the strong contradiction between the derived answer (36) and the provided correct answer (2), it is not possible to provide a valid, step-by-step derivation that reaches 2. The problem statement or the provided answer is likely incorrect. I will proceed with the derived answer of 36, acknowledging the discrepancy.

Summary

The problem asks for the limit of a function as $x$ approaches 2. Direct substitution of $x=2$ into the expression results in the indeterminate form $\frac{0}{0}$. We employ a substitution $t = 3^{x/2}$ to simplify the expression. After performing the substitution and simplifying the numerator and denominator algebraically, we obtain the expression $\frac{(t^2 - 3)(t^2 - 9)}{t - 3}$. Further factorization of the term $t^2 - 9$ as $(t-3)(t+3)$ allows for cancellation of the $(t-3)$ term, leading to the simplified expression $(t^2 - 3)(t + 3)$. Evaluating this simplified expression as $t$ approaches 3 yields $(3^2 - 3)(3 + 3) = (9 - 3)(6) = 6 \times 6 = 36$. This result is also confirmed using L'Hôpital's Rule. There appears to be a discrepancy with the provided correct answer of 2.

Final Answer

Based on standard mathematical derivations, the limit evaluates to 36. There is a discrepancy with the provided correct answer of 2. Assuming the question is stated correctly, the answer is 36.

The final answer is $\boxed{36}$.