Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{1}{x}\log\ (1+7x)$

Answer 1

$\lim_{x \to 0} \: \frac{1}{x} \text{ ln}(1 + 7x)$
Since replacing x with 0 would result in form 0/0,
Apply L'Hopital's Rule.

$\lim_{x \to 0} \: \frac{ \frac{d}{dx} \text{ ln}(1 + 7x)} { \frac{d}{dx} x} \\ \\ = \lim_{x \to 0} \: \frac{7}{1 + 7x} \\ \\ = 7$

Thus,
$\lim_{x \to 0} \: \frac{1}{x} \text{ ln}(1 + 7x) = 7$