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KCET · Maths · Limits

\(\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}\) is equal to

  1. A \(\log _{e} 3\)
  2. B 0
  3. C \(\log _{3} e\)
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(C) \(\log _{3} e\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} \lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}\left(\text { form } \frac{0}{0}\right) \\ &=\lim _{x \rightarrow 0} \frac{\left(\frac{1}{1+x}\right)}{3^{x} \cdot \log _{e} 3} \text { (use L-hospital rule) } \\ &=\frac{\left(\frac{1}{1+0}\right)}{3^{0} \cdot \log _{e} 3} \\ &=\frac{1}{\log _{e} 3}=\log _{3} e \end{aligned}\)