Dude, you used $\LaTeX$. Excellent! I'm very impressed.

It's important to realize that the "let's fiddle around with" part was us experimenting. That was all done under the guise that we were assuming that $|x-2|<\delta$. We were assuming this because our goal was to show that $|f(x)-7|<.1$ **whenever** $|x-2|<\delta$. In other words, we needed to show that **if** $|x-2|<\delta$, **then** $|f(x)-7|<.1$. OK, so if $|x-2|<\delta$, then certainly $3|x-2|<3\delta$. This may look like something wild just happened, but this is the same principle as saying since $4<5$, $3\cdot 4<3\cdot 5$. The reason why $\delta$ must be equal to $.1/3$ (or smaller is fine) is because we want $|f(x)-7|<.1$, but we discovered that $|f(x)-7|<3\delta$ (if $|x-2|<\delta$), and if we just set .1 equal to $3\delta$ and solve for $\delta$, we get $\delta=.1/3$.

I just want to reiterate what I said in class a few times. This stuff is hard and I know that. However, with some effort, everyone can get it. Furthermore, it'll get easier soon.

One last thing…please see my hints for the last two homework problems here. I hope this helps.

Keep up the posting! I love it.