Limit on a function involving natural logarithmns Consider the following question:-
$$\lim_{x \to 3} \frac{\ln(x+4)-\ln(7)}{x-3}$$
I would classify this question as "not knowing where to begin" so I cannot provide my detailed attempt. 
The only clue I was able to notice was that $x+4=x-3+7$ which maybe can be exploited in some way. 
Also, note that the answer is not indeterminate because if we plug $x=3$ into the function, then it yields $0/0$, which means that algebraic manipulation can be used to transform the function into an expression in which two equal terms-one in numerator and other in denominator which both equal $0$ (when $x=3$ is plugged in) can be cancelled so we can remove the $0/0$.
Note: The solution I am seeking is one where L'Hôpital's rule is not used, because it has not been covered in the course I am doing till now.
 A: Simply apply L'Hopital's rule once to get the answer.
$$\lim_{x\to 3}\dfrac{\ln(x+4)-\ln 7}{x-3}=\lim_{x\to 3}\dfrac{\frac{1}{x+4}}{1}=\dfrac{1}{7}$$

Aliter $1$:
Set $h=x-3$. $$\lim_{x\to 3}\dfrac{\ln(x+4)-\ln7}{x-3}=\lim_{h\to 0}\dfrac{\ln(h+7)-\ln7}{h}=\dfrac{\mathrm d}{\mathrm dh}\left(\ln(h+7)\right)\Biggr|_{h\to 0}=\dfrac{1}{7}$$

Aliter $2$:
Set $h=x-3$. And replace $h\mapsto 7h$. Use standard limits.
$$\lim_{h\to 0}\frac{\ln(\frac{h+7}{7})}{h}=\lim_{h\to 0}\frac{\ln(1+h)}{7h}=\frac{1}{7}$$
A: You are working $$y=\frac{\log (x+4)-\log (7)}{x-3}$$ when $x\to 3$.
To make life simpler, let $x=3+t$ to make
$$y=\frac{\log \left(1+\frac{t}{7}\right)}{t}\sim \frac {\frac t7 }t={\frac 17 }$$
A: For fun:
1) Let $x \rightarrow 3^+$.
$\log (\frac{x+4}{7})^{\small{\frac{1}{x-3}}}.$
Set $y= \frac{1}{x-3}.$
Then
$\lim_{y \rightarrow \infty}\log (1+\frac{1}{7y})^y=\log (e^{1/7})=1/7.$
2) Can you do $x \rightarrow 3^-$?
Used: 
$\lim_{x \rightarrow \infty} (1+\frac{a}{x})^x=e^a$, $a$ real.
A: This is figuratively the definition of a derivative in disguise. With a moustache. First substitute h = x - 3
$$\lim_{h \to 0} \frac{\ln(7 + h)-\ln(7)}{h}$$
Where have I seen this before?
$$\frac{dy}{dx} \text{ at } x = 7 =\lim_{h\to 0}\frac{f(7 + h)-f(7)}{h}$$
Therefore your limit must be the derivative of the natural log evaluated at 7.
Your teacher wants you to make the connection between the derivative and the limit.
