Proof for a limit - Epsilon-delta proof $3x^2 +2x +4$

Question

I am trying to figure out what to do when I get stuck trying to prove an epsilon-delta limit.

For example I saw in the textbook with no answer key like this:
$$ \lim_{x \to 3} \; \bigl( 3x^2 + 2x + 4 \bigr) = 37 $$

I usually will try to isolate $x-3$ from $|f(x) -L|$ but I get stuck with something like $|3x^2 + 2x -33|$ which is less than $|3x^2+2x |$ ( assuming you put some bound on $\delta$). but then I get stuck. How would you go about solving this?

Answer 1

With $f(x) = 3x^2 + 2x + 4$ and $L = 37$, you can expand $f(x) - L$ in terms of $x - 3$: $$ 3x^2 + 2x - 33 = 3(x-3)^2 + 20(x-3) $$ so \begin{align} \def\abs#1{\lvert #1 \rvert} \def\Abs#1{\bigl\lvert #1 \bigr\rvert} \Abs{3x^2 + 2x - 33} &= 3\Abs{(x-3)^2 + 20(x-3)} \\ &\leq 3\abs{x-3}^2 + 20\abs{x-3} \\ &= \abs{x-3} \, \bigl( 3\abs{x-3} + 20 \bigr). \end{align}

In order to make this product small, it suffices to make $\abs{x-3} < 1$ so that $$ 3\abs{x-3} + 20 < 23. $$ Then, if you also make $\abs{x-3} < \smash{\frac{\varepsilon}{23}}$, you're in business. In other words, given $\varepsilon > 0$, choose $$ \delta = \min \bigl\{ 1, \tfrac{\varepsilon}{23} \bigr\}. $$ Now, with $\abs{x-3} < \delta$, you get the estimate $$ \def\abs#1{\lvert #1 \rvert} \def\Abs#1{\bigl\lvert #1 \bigr\rvert} \Abs{f(x) - L} = \abs{x-3} \, \bigl( 3\abs{x-3} + 20 \bigr) < \tfrac{\varepsilon}{23} \cdot 23 = \varepsilon. $$