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I have the behavior of this limit: $$ \lim_{x \to 8} \ln (x-8) = -\infty$$ I can't really tell why it's happening, and i have to understand it, i think that's because what's in the ln is tending to $0$ so i can also assume that as well? $$\lim_{ x \to 0} \ln x = -\infty$$ will be happy to understand this concept it purely and not just by rules given to me in school.

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First note that $\ln(x-8)$ is defined only if $x-8 \gt 0 \iff x \gt 8\,$. This means that $x$ can only tend to $8$ from the right, and it is good practice to make that obvious by writing the limit as $x \to 8^+\,$.

Then, to formalize the intuition about the relation to $\lim_{x \to 0^+} \ln (x)\,$, let $y=x-8\,$ so that $y \to 0^+$ when $x \to 8^+$. It follows that:

$$\lim_{x \to 8+} \ln (x-8) = \lim_{y \to 0^+} \ln (y) = -\infty$$

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    very helpful, could understand it with ease, thank you. – 2017-01-10
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Yes, you are correct, although technically it should be a limit from the right, not a two-sided limit i.e. $$ \lim_{x\rightarrow 8^+} = -\infty.$$ In general, this method of looking for the limit of what's inside the function (in this case $\lim_{x\rightarrow 8^+} x-8 = 0^+$) and then looking what the function does at that limit (i.e. $\lim_{y\rightarrow 0^+}\ln(y) = -\infty$) is a good way to reason about limits.

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    i know that I can't tend to it from the left side because it's undefined in $x< 0$ but can't it tend only from $0 – 2017-01-09
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    @Ozk not sure I understand. $\ln(y)$ is undefined for $y<0$ (at least as a real-valued function) so the limit $\lim_{y\rightarrow 0^-}\ln(y)$ is undefined. Thus $\lim_{x\rightarrow 8^-}\ln(x-8)$ is undefined, as is $\ln(x-8)$ for all $x\le 8.$ – 2017-01-10