KCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}\) is equal to
- A \(\log _{e} 3\)
- B 0
- C \(\log _{3} e\)
- D 1
Answer & Solution
Correct Answer
(C) \(\log _{3} e\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}\left(\text { form } \frac{0}{0}\right) \\ &=\lim _{x \rightarrow 0} \frac{\left(\frac{1}{1+x}\right)}{3^{x} \cdot \log _{e} 3} \text { (use L-hospital rule) } \\ &=\frac{\left(\frac{1}{1+0}\right)}{3^{0} \cdot \log _{e} 3} \\ &=\frac{1}{\log _{e} 3}=\log _{3} e \end{aligned}\)
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