Limit proof by definition I have the following limit:
$$\lim_{x \to 6} \frac{x+1}{x-5} = 7$$
How do I prove this equation by the famous definition of the limit? (the one includes delta and epsilon).
I know how to prove simple limits like this one: $$\\\lim_{x \to 6} x-5 = 1 
\\ |x - x_0| < \delta \implies |x - 6| < \delta 
\\ |f(x) -L| < \epsilon \implies |x-5 -1| < \epsilon  \implies |x - 6| < \epsilon  
\\ \implies \delta = \epsilon$$
But in the first example I asked about I have:
$$|x-6| < \delta$$ $$|\frac{x+1}{x-5} -7| < \epsilon \implies |\frac{36-6x}{x-5}| < \epsilon$$
don't really know how to proceed, thanks in advance :)

We have $$\bigg|\frac{x+1}{x-5}-7\bigg|=\bigg|\frac{x+1-7x+35}{x-5}\bigg|=\bigg|\frac{-6x+36}{x-5}\bigg|=6\cdot\bigg|\frac{x-6}{x-5}\bigg|=6\cdot\bigg|\frac{1}{x-5}\bigg|\cdot|x-6|$$
Assume that $|x-6|<\frac{1}{2}$. Then we get 
$$-\frac{1}{2}<x-6<\frac{1}{2}$$ and so we get
$$-\frac{1}{2}+1<x-6+1<\frac{1}{2}+1$$ that is, we get
$$0<\frac{1}{2}<x-5<\frac{3}{2}$$ that is, we get
$$0<\frac{1}{x-5}<2.$$
Let $\epsilon>0$.
Define $$\delta=\min\bigg\{\frac{1}{2},\frac{\epsilon}{12}\bigg\}.$$ Then $\delta\leq\frac{1}{2}$ and $\delta\leq\frac{\epsilon}{12}$. Hence, if $0<|x-6|<\delta$ then we get
$$\begin{align}\bigg|\frac{x+1}{x-5}-7\bigg|&=6\cdot\bigg|\frac{1}{x-5}\bigg|\cdot|x-6|\\
&=6\cdot\frac{1}{x-5}\cdot|x-6|\\
&<12\cdot\delta\leq \epsilon.
\end{align}$$
