AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}\) is equal to
- A \(\left(\log _e 2\right) \log _e 3\)
- B \(\log _{\mathrm{e}} 5\)
- C \(\log _{\mathrm{e}} 6\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(\left(\log _e 2\right) \log _e 3\)
Step-by-step Solution
Detailed explanation
Let, \(l=\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}=\lim _{x \rightarrow 0} \frac{3^x \cdot 2^x-3^x-2^x+1}{x^2}\)…
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