Finding Basic Limits For the function given by $f(x) = |x − 2| + 42$ and $x → 2$, make a guess of the limit, and then prove your
  assertion using the formal definition of a limit.


This was confusing; please help.

My guess is $f(x) = 42$ as $x$ approaches $2$.
Claim: $\displaystyle \lim_{x \to 2} f(x) = 42$.

Proof: Let $\epsilon &gt; 0$ be given, choose $\delta = \epsilon$. If $0 &lt; |x-2| &lt; \delta \Rightarrow |f(x)-42| = ||x-2|+42-42| = |x-2| &lt; \delta = \epsilon$, hence the result.
A function can have a limit so long as for $f(x)=L$ such that there is a distance $\delta$ to the left and right of $a$ such that there correspond to $x$ values that produce $y$ outputs that are $\epsilon$ values on the top and bottom of the limit.

$|x-a|&gt;\delta$ such that $|f(x)-L| &gt; \epsilon$

$f(2)=42$ as $x$ approaches 
So....
$|x-2|&gt;\delta$
say $\epsilon=0.01$, and we have values $y=42.01$ and $y=41.99$ casing $42$. 

If $42.01=|x-2|+42$, then $2.01=x$.

If $41.99=|x-2|+42$, then $-0.99+2=1.99$.

If $2+\delta=2.01$, $2-\delta=1.99$, then $\delta=0.01$.

Hence there exists these two $\delta$ values such that there is also an $\epsilon$ distance between $42$.