You're missing symbols in both of your expressions, but considering the first limit has a value of 10, I suspect you meant to write
(which is true) but unfortunately I am nowhere near as confident about what the second one is supposed to say. So one proof will have to do, unless you come around to editing your question.
The claim,
is to say that, for any given <em>ε</em> > 0, we can find <em>δ</em> (a number that depends on <em>ε</em>) such that whenever |<em>x</em> - 2| < <em>δ</em>, this ensures that |(3<em>x</em> + 4) - 10| < <em>ε</em>.
Roughly speaking: if <em>x</em> is close enough to 2, this translates to <em>f(x)</em> = 3<em>x</em> + 4 being close enough to 10. It's our job to figure out how close <em>x</em> needs to be to 2 in order that <em>f(x)</em> is close enough to 10, where the closeness to 10 is some given threshold.
We want to arrive at the inequality,
|(3<em>x</em> + 4) - 10| < <em>ε</em>
so suppose we work backwards. With some simplification and rewriting, we have
|3<em>x</em> - 6| = |3 (<em>x</em> - 2)| = |3| |<em>x</em> - 2| = 3 |<em>x</em> - 2| < <em>ε</em>
and so
|<em>x</em> - 2| < <em>ε</em>/3
which suggests that we should pick <em>δ</em> = <em>ε</em>/3.
Now for the proof itself:
Let <em>ε</em> > 0 be given, and let <em>δ</em> = <em>ε</em>/3. Then
|<em>x</em> - 2| < <em>δ</em> = <em>ε</em>/3
3 |<em>x</em> - 2| < <em>ε</em>
|3<em>x</em> - 6| < <em>ε</em>
|(3<em>x</em> + 4) - 10| < <em>ε</em>
and this completes the proof of the limit. QED