MHT CET · Maths · Limits
The value of \(\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}\) is
- A \(\frac{(\log 3)(\log 5)}{2}\)
- B \(2(\log 3)(\log 5)\)
- C \(\frac{\log 3+\log 5}{2}\)
- D None of these
Answer & Solution
Correct Answer
(A) \(\frac{(\log 3)(\log 5)}{2}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}\)
\(=\lim _{x \rightarrow 0} \frac{\left(3^{x}-1\right)\left(5^{x}-1\right)}{1-1+2 \sin ^{2} x}\)
\(\quad=\lim _{x \rightarrow 0}\left(\frac{3^{x}-1}{x}\right)\left(\frac{5^{x}-1}{x}\right)\left(\frac{x^{2}}{2 \sin ^{2} x}\right)\)
\(=\frac{1}{2}(\log 3)(\log 5) \cdot 1\)
\(\quad\left[\because \lim _{x \rightarrow 0}\left(\frac{a^{x}-b^{x}}{x}\right)=\log \left(\frac{a}{b}\right)\right]\)
\(=\lim _{x \rightarrow 0} \frac{\left(3^{x}-1\right)\left(5^{x}-1\right)}{1-1+2 \sin ^{2} x}\)
\(\quad=\lim _{x \rightarrow 0}\left(\frac{3^{x}-1}{x}\right)\left(\frac{5^{x}-1}{x}\right)\left(\frac{x^{2}}{2 \sin ^{2} x}\right)\)
\(=\frac{1}{2}(\log 3)(\log 5) \cdot 1\)
\(\quad\left[\because \lim _{x \rightarrow 0}\left(\frac{a^{x}-b^{x}}{x}\right)=\log \left(\frac{a}{b}\right)\right]\)
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