Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} \frac{(x+4)}{4}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (\frac{x}{4}+1)$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (1 + \frac{x}{4})$$
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1}{x}$$
Then multiply the power, by 4.
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1.4}{x.4}$$
$$\ln \begin{bmatrix}\lim_{x\to 0} (1 + \frac{x}{4})^\frac{4}{x} \end{bmatrix}^\frac{1}{4} $$
$$ \ln \phantom{2}\mathcal e^\frac{1}{4} = \frac{1}{4}$$
Until here is fine, but someone's telling me that it can be made like this:
$$\lim_{x\to 0} \frac{\ln(x) + \ln(4) -\ln(4)}{x}$$
$$\lim_{x\to 0} \frac{\ln(x)}{x} = 1$$
Alright, it uses logarithmic property to separate the expression, but i said it can't be, because of the indetermination, and $\frac{1}{4} $ is closer of $0$ than $1$. There's another reasonable explanation for this ? Why it can't be or when can it ?
And, there's a trick to way out quickly with this limit, keeping L'Hôspital  aside ?

Consider rewriting the expression as
$\dfrac{\ln(x + 4) - \ln(4)}{x} = \dfrac{\ln\Big(\dfrac{x + 4}{4}\Big)}{x} = \ln\Big(\frac{x}{4} + 1\Big)^{1/x} = \frac{1}{4}\ln\Big(\frac{1}{4/x} + 1\Big)^{4/x}$
Then, we find
$\lim_{x \rightarrow 0} \dfrac{\ln(x + 4) - \ln(4)}{x} = \frac{1}{4}\ln\lim_{x \rightarrow 0}\Big(\frac{1}{4/x} + 1\Big)^{4/x} = \frac{1}{4}\ln(e) = \frac{1}{4}$
