Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c} f(x) = f(c)$.
• Standard Limits:
  • $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$
  • $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
• Algebraic Manipulation: Factoring and rationalization techniques are often useful in evaluating limits.

Step-by-Step Solution

Step 1: Understanding the Condition for Continuity The problem states that the function $f(x)$ is continuous at $x=0$. For a function to be continuous at a point, the limit of the function as $x$ approaches that point must exist and be equal to the function's value at that point. Therefore, we must have: $\lim_{x \to 0} f(x) = f(0)$ Given the function definition: $f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ a \log_e 2 \log_e 3, & x = 0 \end{cases}$ So, the condition becomes: $\lim_{x \to 0} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = a \log_e 2 \log_e 3$

Step 2: Evaluating the Numerator of the Limit Let's focus on the numerator of the limit: $72^x - 9^x - 8^x + 1$. We can factor this expression. Notice that $72 = 9 \times 8$. $72^x - 9^x - 8^x + 1 = (9 \times 8)^x - 9^x - 8^x + 1$ $= 9^x 8^x - 9^x - 8^x + 1$ We can group terms to factor: $= 9^x (8^x - 1) - 1 (8^x - 1)$ $= (9^x - 1)(8^x - 1)$ So, the limit expression becomes: $\lim_{x \to 0} \frac{(9^x - 1)(8^x - 1)}{\sqrt{2} - \sqrt{1 + \cos x}}$

Step 3: Evaluating the Denominator of the Limit and Rationalizing The denominator is $\sqrt{2} - \sqrt{1 + \cos x}$. As $x \to 0$, $\cos x \to \cos 0 = 1$. So the denominator approaches $\sqrt{2} - \sqrt{1+1} = \sqrt{2} - \sqrt{2} = 0$. This confirms we have an indeterminate form $0/0$. To simplify the denominator, we can rationalize it by multiplying the numerator and denominator by the conjugate: $\sqrt{2} + \sqrt{1 + \cos x}$. $\frac{(9^x - 1)(8^x - 1)}{\sqrt{2} - \sqrt{1 + \cos x}} \times \frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$ $= \frac{(9^x - 1)(8^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{(\sqrt{2})^2 - (\sqrt{1 + \cos x})^2}$ $= \frac{(9^x - 1)(8^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{2 - (1 + \cos x)}$ $= \frac{(9^x - 1)(8^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$

Step 4: Rearranging the Limit Expression for Standard Forms Now, let's rewrite the limit expression using the result from Step 3: $\lim_{x \to 0} \frac{(9^x - 1)(8^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$ We want to use the standard limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$. We can rearrange the terms: $\lim_{x \to 0} \left( \frac{9^x - 1}{x} \right) \left( \frac{8^x - 1}{x} \right) \left( \frac{x^2}{1 - \cos x} \right) (\sqrt{2} + \sqrt{1 + \cos x})$ Note that we multiplied and divided by $x$ for the first two terms and by $x^2$ for the third term, which is equivalent to multiplying the entire expression by $x \cdot x / x^2 = 1$.

Step 5: Evaluating the Limit Now, we can evaluate each part of the rearranged limit:

• $\lim_{x \to 0} \frac{9^x - 1}{x} = \ln 9$
• $\lim_{x \to 0} \frac{8^x - 1}{x} = \ln 8$
• $\lim_{x \to 0} \frac{x^2}{1 - \cos x} = \frac{1}{\lim_{x \to 0} \frac{1 - \cos x}{x^2}} = \frac{1}{1/2} = 2$
• $\lim_{x \to 0} (\sqrt{2} + \sqrt{1 + \cos x}) = \sqrt{2} + \sqrt{1 + \cos 0} = \sqrt{2} + \sqrt{1+1} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$

Multiplying these results together: $\lim_{x \to 0} f(x) = (\ln 9) (\ln 8) (2) (2\sqrt{2})$ $= 4\sqrt{2} \ln 9 \ln 8$ We can simplify $\ln 9$ and $\ln 8$ using logarithm properties ($\ln a^b = b \ln a$):

• $\ln 9 = \ln 3^2 = 2 \ln 3$
• $\ln 8 = \ln 2^3 = 3 \ln 2$

Substituting these back: $\lim_{x \to 0} f(x) = 4\sqrt{2} (2 \ln 3) (3 \ln 2)$ $= 4\sqrt{2} \cdot 6 \ln 2 \ln 3$ $= 24\sqrt{2} \ln 2 \ln 3$

Step 6: Equating the Limit to $f(0)$ and Solving for $a$ From Step 1, we know that $\lim_{x \to 0} f(x) = f(0)$. We found $\lim_{x \to 0} f(x) = 24\sqrt{2} \ln 2 \ln 3$. And we are given $f(0) = a \log_e 2 \log_e 3$. Equating these two: $24\sqrt{2} \ln 2 \ln 3 = a \ln 2 \ln 3$ Since $\ln 2 \neq 0$ and $\ln 3 \neq 0$, we can divide both sides by $\ln 2 \ln 3$: $a = 24\sqrt{2}$

Step 7: Calculating $a^2$ The question asks for the value of $a^2$. $a^2 = (24\sqrt{2})^2$ $a^2 = 24^2 \times (\sqrt{2})^2$ $a^2 = 576 \times 2$ $a^2 = 1152$

Common Mistakes & Tips

• Incorrectly Factoring the Numerator: Ensure that $72^x - 9^x - 8^x + 1$ is correctly factored into $(9^x - 1)(8^x - 1)$. A common error is to assume $72^x = 9^x 8^x$ which is true.
• Handling the Denominator: Rationalizing the denominator is crucial. Forgetting to multiply both the numerator and denominator by the conjugate, or making errors in the algebraic expansion of $(\sqrt{2} - \sqrt{1 + \cos x})(\sqrt{2} + \sqrt{1 + \cos x})$, can lead to incorrect results.
• Applying Standard Limits: Ensure that the expression is manipulated to match the standard limit forms. For instance, dividing by $x$ and $x^2$ appropriately when using $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

Summary

The problem requires finding the value of $a^2$ given that the function $f(x)$ is continuous at $x=0$. Continuity implies that the limit of the function as $x$ approaches 0 must equal the function's value at $x=0$. We evaluated the limit by first factoring the numerator, rationalizing the denominator, and then strategically rearranging the expression to utilize standard limit formulas like $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$. After calculating the limit, we equated it to $f(0)$ to find the value of $a$, and subsequently calculated $a^2$.

The final answer is $\boxed{1152}$.