Problem with understanding a limit proof technique So recently we studied about the epsilon delta definition of a limit in calculus.
I have my exercise and just wrote out a proof 
without understanding what I wrote at all
just did what my professor did in class 1 to 1. 
Can you please explain me how this can be a proof of a limit and if it's wrong how should I approach tese questions ? 
I get it for the linear equations case ... 
but in cases like this when I have another absolute value that contains x  besides my delta expression I  just get confused. 
I restrict delta to be less than 1 ( this I get i need to approach my x not get far from it)
But the last part when I choose delta to be the minimum of 2 numbers - it conjunction how a conjunction between these two can imply something ? 
Thanks for helping here is my proof as an example (sorry I dont know how to type mathematics in this website)
 
 A: The definition of limit is a theoretical test to check whether a given number $L$ is a limit of a function $f(x)$ or not. It is not a practical technique to evaluate a limit.
The current question therefore mentions to guess the limit and prove that the guessed number is a limit according to definitions. I hope you have understood the part of the proof upto the following inequality $$\left|\frac{x - 4}{x - 2}\right| < 6\epsilon\tag{1}$$ Our task now is to find a suitable number $\delta > 0$ such that if $0 < |x - 2| < \delta$, then the above inequality $(1)$ holds. The idea is to find an expression (say $g(x)$) which is simpler than the LHS of $(1)$ and at the same time greater than the LHS of $(1)$. And then if we can somehow guarantee that $g(x) < 6\epsilon$ then inequality $(1)$ will be satisfied automatically.
So our task of ensuring the inequality $(1)$ breaks down into two steps:


*

*Finding a simple expression $g(x)$ such that LHS of $(1)$ is less than $g(x)$.

*Ensure that $g(x) < 6\epsilon$


Both these goals will clearly be possible for certain values of $x$ only and not for all values of $x$. We want them to be valid for those $x$ under the constraint $0 < |x - 2| < \delta$. Thus we need to find $\delta$ such that both the goals are achieved. For first goal $\delta \leq 1$ is sufficient. Assuming that the first goal is achieved, we further observe that $\delta \leq 6\epsilon$ in order to achieve second goal. Thus in order to achieve both the goals simultaneously we see that the value of $\delta$ should be such that $\delta \leq 1$ and $\delta \leq 6\epsilon$. Hence we choose $\delta = \min(1, 6\epsilon)$ and then both our goals are achieved simultaneously for $0 < |x - 2| < \delta$.
