MHT CET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=\)
- A \(8 \sqrt{5} \log 3\)
- B \(16 \sqrt{5} \log 3\)
- C \(8 \sqrt{5}(\log 3)^2\)
- D \(\sqrt{5}(\log 3)^2\)
Answer & Solution
Correct Answer
(C) \(8 \sqrt{5}(\log 3)^2\)
Step-by-step Solution
Detailed explanation
\( \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x-1\right)}{5-(4+\cos x)}\{\sqrt{5}~+\) \(\sqrt{4+\cos x}\} \)
\( =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x+1\right)}{1-\cos x} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\} \)
\( =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x-1\right)}{2 \sin ^2\left(\frac{x}{2}\right)} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\} \)
\( =\lim _{x \rightarrow 0} \frac{\left(\frac{9^x-1}{x}\right)\left(\frac{3^x-1}{x}\right)}{2 \times \frac{\sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{1}{4}} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\}\)
\(\begin{aligned} & =\frac{\log 9 \cdot \log 3}{\frac{1}{2}}\{\sqrt{5}+\sqrt{5}\} \\ & =2 \log 9 \cdot \log 3 \cdot(2 \sqrt{5}) \\ & =8 \sqrt{5}(\log 3)^2\end{aligned}\)
\( =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x+1\right)}{1-\cos x} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\} \)
\( =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x-1\right)}{2 \sin ^2\left(\frac{x}{2}\right)} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\} \)
\( =\lim _{x \rightarrow 0} \frac{\left(\frac{9^x-1}{x}\right)\left(\frac{3^x-1}{x}\right)}{2 \times \frac{\sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{1}{4}} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\}\)
\(\begin{aligned} & =\frac{\log 9 \cdot \log 3}{\frac{1}{2}}\{\sqrt{5}+\sqrt{5}\} \\ & =2 \log 9 \cdot \log 3 \cdot(2 \sqrt{5}) \\ & =8 \sqrt{5}(\log 3)^2\end{aligned}\)
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