Question

$\lim_{x \to 0} \frac{63^x - 9^x - 7^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = \dots$

• A. $4\sqrt{2} / (\log 7 \cdot \log 9)$
• B. $4\sqrt{2} \log 7 \cdot \log 9$
• C. $4\sqrt{2} \log 63$
• D. $(\log 7 \cdot \log 9) / 4\sqrt{2}$

Hint:

To avoid tedious conjugation steps with cosine limits, memorize the Maclaurin series approximation for small angles: $1 - \cos(kx) \approx \frac{(kx)^2}{2}$. For $1 - \cos(x/2)$, it instantly becomes $\frac{(x/2)^2}{2} = \frac{x^2}{8}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must evaluate a $0/0$ limit. The numerator is a mixture of exponential terms, and the denominator contains square roots and a trigonometric function.

Step 2: Key Formula or Approach:
1. Factor the numerator by grouping to expose terms of the form $(a^x - 1)$.
2. Use the standard limit identity: $\lim_{x \to 0} \frac{a^x - 1}{x} = \log_e a$.
3. Use the half-angle formula on the denominator: $1 + \cos x = 2\cos^2(x/2)$.
4. Use the small angle approximation for cosine: $1 - \cos\theta \approx \frac{\theta^2}{2}$.

Step 3: Detailed Explanation:
Let's restructure the numerator:
Since $63 = 9 \times 7$, $63^x = 9^x \cdot 7^x$.
$$9^x \cdot 7^x - 9^x - 7^x + 1 = 9^x(7^x - 1) - 1(7^x - 1) = (9^x - 1)(7^x - 1)$$
Now let's restructure the denominator:
$$\sqrt{2} - \sqrt{1 + \cos x} = \sqrt{2} - \sqrt{2\cos^2(x/2)} = \sqrt{2} - \sqrt{2}\cos(x/2) = \sqrt{2}\left(1 - \cos\frac{x}{2}\right)$$
Substitute these back into the limit:
$$L = \lim_{x \to 0} \frac{(9^x - 1)(7^x - 1)}{\sqrt{2}(1 - \cos\frac{x}{2})}$$
Divide the numerator and denominator by $x^2$:
$$L = \lim_{x \to 0} \frac{ \left(\frac{9^x - 1}{x}\right) \left(\frac{7^x - 1}{x}\right) }{ \sqrt{2} \left( \frac{1 - \cos(x/2)}{x^2} \right) }$$
The limit of the numerator pieces:
$$\lim_{x \to 0} \frac{9^x - 1}{x} = \log 9 \quad \text{and} \quad \lim_{x \to 0} \frac{7^x - 1}{x} = \log 7$$
For the denominator, multiply and divide by 4 to create the standard format:
$$\lim_{x \to 0} \frac{1 - \cos(x/2)}{x^2} = \lim_{x \to 0} \frac{1}{4} \frac{1 - \cos(x/2)}{(x/2)^2} = \frac{1}{4} \left(\frac{1}{2}\right) = \frac{1}{8}$$
Substitute all these evaluated components back together:
$$L = \frac{ (\log 9)(\log 7) }{ \sqrt{2} \left( \frac{1}{8} \right) } = \frac{8 \log 9 \log 7}{\sqrt{2}}$$
Rationalize by multiplying top and bottom by $\sqrt{2}$:
$$L = \frac{8\sqrt{2} \log 9 \log 7}{2} = 4\sqrt{2} \log 7 \log 9$$

Step 4: Final Answer:
The limit is $4\sqrt{2} \log 7 \log 9$, matching option (b).