MHT CET · Maths · Limits
If \(\mathrm{f}(x)=\left(\frac{2^{x}-1}{1-3^{x}}\right)\), for \(\mathrm{x} \neq 0\) is continuous at \(x=0\), then \(\mathrm{f}(0)=\)
- A \(\cdot \log 3\)
- B \(\frac{-(\log 2)}{(\log 3)}\)
- C \(\frac{(\log 2)}{(\log 3)}\)
- D \(-\log 2\)
Answer & Solution
Correct Answer
(B) \(\frac{-(\log 2)}{(\log 3)}\)
Step-by-step Solution
Detailed explanation
\(f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}\)
\(
f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}
\)
\(
f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}
\)
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