Q.Evaluate $\displaystyle\lim_{x\to0}\frac{\ln(1+7x)}{x}$.

This matches $\dfrac{\ln(1+kx)}{x}$ with $k=7$, which equals $k$ directly.\n> [!TLDR]\n> Scaled logarithmic-limit formula with $k=7$.\n …

Direct substitution gives $\dfrac{\ln(1)}{0}=\dfrac00$ — indeterminate, so the standard logarithmic limit (§5) applies.

Substitute $u=7x$ (so $u\to0$ as $x\to0$, and $x=u/7$):

$\lim_{x\to0}\frac{\ln(1+7x)}{x} = \lim_{u\to0}\frac{\ln(1+u)}{u/7} = 7\cdot\lim_{u\to0}\frac{\ln(1+u)}{u} = 7(1) = 7.$ …

Numeric verification (shown above): $x=0.001$ gives a ratio close to $7$, and the ratio approaches $7$ more closely as $x$ shrinks further, c …

Students sometimes forget the argument must be exactly $(1+kx)$, not $(kx)$ alone — $\ln(kx)$ (without the $+1$) is NOT covered by this standard limit and does not even approach a finite limit of this si …