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 ?
 A: Alternatively, you could also observe that the limit is just the derivative of $x \to ln (x) $ at $x=4$.
A: 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
$\lim_{x \rightarrow 0} \dfrac{\ln(x + 4) - \ln(4)}{x} = \frac{1}{4}\ln\lim_{x \rightarrow 0}\Big(\frac{1}{4/x} + 1\Big)^{4/x} = \frac{1}{4}\ln(e) = \frac{1}{4}$
A: $\ln(x+4)\neq \ln(x)+\ln(4)$ and additionally $\lim_{x\to 0} \frac{\ln(x)}{x}\neq 1$
The first one is correct.
A: Yet another method —Taylor expansion (to first order):
We use that $\ln(1+u) = u + o(u)$ when $u\to 0$.
Then $\ln(x+4) = \ln\left(1+\frac{x}{4}\right) + \ln 4 = \frac{x}{4} + o(x) + \ln 4$, and
$$
\frac{\ln(x+4)- \ln 4}{x} = \frac{\frac{x}{4} + o(x)}{x} = \frac{1}{4} + o(1) \xrightarrow[x\to0]{} \frac{1}{4}.
$$
A: You want
$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}
$.
One of the definitions 
of $\ln$ is that
$\ln(x)
=\int_1^x \frac{dt}{t}
$.
Therefore
$\ln(x+4)-\ln(4)
=\int_4^{x+4} \frac{dt}{t}
$.
If
$x > 0$
then,
since
$\frac14 
\ge \frac{1}{t}
\ge \frac1{x+4}
$
for
$4 \le t \le 4+x$,
$\frac{x}{4} 
\ge \int_4^{x+4}\frac{dt}{t}
\ge \frac{x}{x+4}
$,
so that
$\frac{1}{4} 
\ge \frac1{x}\int_4^{x+4}\frac{dt}{t}
\ge \frac{1}{x+4}
$.
Therefore
$\lim_{x \to 0^+}\frac1{x}\int_4^{x+4}\frac{dt}{t}
=\frac14
$.
We can similarly show that
$\lim_{x \to 0^-}\frac1{x}\int_4^{x+4}\frac{dt}{t}
=\frac14
$.
Therefore the limit is
$\frac14$.
A: Whenever you are dealing with limit of a function you must know certain properties of the function involved. Especially the focus should be on the analytic (and not merely algebraic) properties of the function. More importantly for every kind of elementary function there is a standard limit formula which can be proved by using an appropriate definition of the function involved and the theorems on limits.
The following is a list of such formulas:


*

*Algebraic functions: $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$

*Trigonometric Functions: $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$

*Logarithmic Functions: $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\, \lim_{x \to \infty}\frac{\log x}{x} = 0$$

*Exponential Functions: $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\, \lim_{x \to \infty}\frac{x}{e^{x}} = 0$$


Note that each of the formulas above either deals with the indeterminate form $0/0$ or $\infty/\infty$. Some may prefer to think of these formulas (the one which deal with form $0/0$) to be equivalent to the derivative formulas for the functions involved, but since I learned calculus via the route of limits, continuity, derivative, integral I prefer to assume the above formulas to be fundamental and the derivatives being evaluated by making use of the above formulas (and yes each of them has a proof without using anything about derivatives so that no circularity is involved in the use of these formulas). Many people might prefer to add more formulas in the above list but the above is sufficient. Moreover since exponential and logarithmic functions are inverse to each other, formulas related to only one of them is sufficient, but I have added them to avoid hassle of converting between direct and inverse functions.

Your question deals with logarithm function and it is expected that one should make use of the standard limit formula involving $\log$ function. Thus we can proceed as follows:
\begin{align}
L &= \lim_{x \to 0}\frac{\log(x + 4) - \log 4}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(\dfrac{x + 4}{4}\right)}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{x}{4}\right)}{\dfrac{x}{4}}\cdot\frac{1}{4}\notag\\
&= \frac{1}{4}\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = x/4)\notag\\
&= \frac{1}{4}\notag
\end{align}
The use of L'Hospital's Rule in such problems is not justified because of the circularity involved. To apply L'Hospital's Rule you need to know the derivative of $\log x$ and if you already know that derivative of $\log x$ is $1/x$ then you can clearly see that the above limit is the derivative of $\log x$ at $x = 4$ and hence it is $1/4$ and thus there is no need to go for L'Hospital's Rule.
A: 
Note that in THIS ANSWER, I showed using on the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$
for $x>0$.

Writing $\log(x+4)-\log(4)=\log\left(1+\frac x4\right)$ and appealing to $(1)$, with $x$ replaced by $1+x/4$, we find that
$$\frac{1/4}{1+x/4}\le \frac{\log\left(1+\frac x4\right)}{x}\le \frac 14 \tag 2$$
whence application of the squeeze theorem yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0} \frac{\log\left(1+\frac x4\right)}{x}=\frac14}$$
And we are done!
