_ x 0 _e(1+x) 3^x-1 = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{3^x-1}=$

A
$\log _e 3$
B
$0$
C
1
✓
$\log _3 e$

✓

Answer

Correct option: D.

$\log _3 e$

(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{3^x-1}=\lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{3^x \log _e 3}=\frac{1}{\log _e 3}=\log _3 e$
Alternate method:
$\lim _{x \rightarrow 0} \frac{\frac{\log _e(1+x)}{x}}{\frac{3^x-1}{x}}=\frac{\log e }{\log 3}=\log _3 e$

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