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 Concept:
Factorize the numerator and rationalize the denominator. Note that $63^x = 9^x \cdot 7^x$.

Step 2: Formula Application:
Numerator: $9^x(7^x - 1) - 1(7^x - 1) = (9^x - 1)(7^x - 1)$. Denominator: Rationalize by multiplying by $\sqrt{2} + \sqrt{1 + \cos x}$. $\sqrt{2}^2 - (1 + \cos x) = 2 - 1 - \cos x = 1 - \cos x$.

Step 3: Explanation:
The limit becomes: $\lim_{x \to 0} \frac{(9^x - 1)(7^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$. As $x \to 0$, $(\sqrt{2} + \sqrt{1 + \cos 0}) = 2\sqrt{2}$. Use standard limits: $\frac{a^x - 1}{x} \to \log a$ and $\frac{1 - \cos x}{x^2} \to \frac{1}{2}$. $L = \frac{(\frac{9^x-1}{x} \cdot x)(\frac{7^x-1}{x} \cdot x) \cdot 2\sqrt{2}}{\frac{1-\cos x}{x^2} \cdot x^2} = \frac{\log 9 \cdot \log 7 \cdot 2\sqrt{2}}{1/2} = 4\sqrt{2} \log 9 \log 7$.

Step 4: Final Answer:
The limit is $4\sqrt{2} \log 7 \log 9$.