limx→0 2log⁡(1+x)−log⁡(1+2x)x2 is equal to

A
0
B
1
C
$- 1$
D
2

Solution:

$lim_{x \to 0} \frac{2 \log (1 + x) - \log (1 + 2 x)}{x^{2}}$

$\begin{array}{l} = lim_{x \to 0} \frac{\log \{\frac{(1 + x)^{2}}{1 + 2 x}\}}{x^{2}} \\ = lim_{x \to 0} \frac{\log (1 + \frac{x^{2}}{1 + 2 x})}{\frac{x^{2}}{1 + 2 x}} \times \frac{1}{1 + 2 x} = 1 \end{array}$

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