ₓ arrow 0} {63^{x}-9^{x}-7^{x}+1}{{2}-{1+ x}}=

Correct Answer is : 42log⁡7⋅log⁡94 {2} 7 · 942log7⋅log9

Correct Answer is : 42log⁡7⋅log⁡94 {2} 7 · 942log7⋅log9

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MHTCET 25 Apr 2025 Shift 2 Paper

Question : 104 of 150

Marks: +1, -0

\lim _{x \rightarrow 0} \frac{63^{x}-9^{x}-7^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}=\ldots \ldots

Solution:

Concept:

Use factorization and rationalization, then apply standard limits

\lim_{x \to 0} \frac{a^x - 1}{x} = \log a

and

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

Explanation:

Rewrite

63^x

9^x \cdot 7^x

and rationalize the denominator.

\lim_{x \to 0} \frac{9^x 7^x - 9^x - 7^x + 1}{\sqrt{2} - \sqrt{1+\cos x}} \times \frac{\sqrt{2} + \sqrt{1+\cos x}}{\sqrt{2} + \sqrt{1+\cos x}}

Factor the numerator and simplify the denominator:

= \lim_{x \to 0} \frac{(7^x - 1)(9^x - 1)(\sqrt{2} + \sqrt{1+\cos x})}{1 - \cos x}

Now split into standard limit forms:

= \lim_{x \to 0} \left( \frac{7^x - 1}{x} \cdot \frac{9^x - 1}{x} \cdot \frac{x^2}{1 - \cos x} \cdot (\sqrt{2} + \sqrt{1+\cos x}) \right)

Evaluate each factor:

\frac{7^x - 1}{x} \to \log 7

\frac{9^x - 1}{x} \to \log 9

\frac{x^2}{1 - \cos x} \to 2

, and

\sqrt{2} + \sqrt{1+\cos x} \to 2\sqrt{2}

Therefore, the limit is:

\log 7 \cdot \log 9 \cdot 2 \cdot 2\sqrt{2} = 4\sqrt{2} \log 7 \cdot \log 9

Answer:

Option B:

4 \sqrt{2} \log 7 \cdot \log 9

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