So it appears that for these $x$, $P(x)$ is relatively close to $5$, within $1$ of $5$.$2\cdot(2.5) \lt 2\cdot x\lt 2\cdot(3.5)$ and
$2\cdot(2.5)-1 \lt 2\cdot x-1\lt 2\cdot(3.5)-1$ so
$4 \lt P(x) \lt 6$ or
$-1 \lt P(x)-5 \lt 1$.
Thus $| P(x) -5| \lt 1$.
$2\cdot(3-\delta^*) \ \lt 2\cdot x\lt 2\ \cdot(3+\delta^*)$ and
$2\cdot(3-\delta^*)-1 \lt 2\cdot x-1\lt 2\cdot(3+\delta^*)-1$ so
$5-2\delta^* \ \ \lt P(x) \ \ \lt 5 + 2\delta^*$ and
$\ \ -2\delta^* \ \lt P(x)-5 \ \lt 2\delta^*$.
Thus $|P(x)-5| \lt 2\delta^*$
i. $L = L^*$. [Limits are unique.] ii. $ \lim\limits_{x \rightarrow a} \ P(x) \pm Q(x)= L \pm M $. iii. $ \lim\limits_{x \rightarrow a} \ c \cdot P(x) = c\cdot L$. |
iv. $ \lim\limits_{x \rightarrow a} \ P(x)\cdot Q(x) = L\cdot M$. v. If $M \ne 0$ then $ \lim\limits_{x \rightarrow a} \ \frac {P(x)}{Q(x)} = \frac LM$. vi. If $R(x)$ is between $P(x)$ and $Q(x)$ for the deleted interval and $M = L$ then $ \lim\limits_{x \rightarrow a} \ R(x) = L$ |
i. $S(x) = P(x) \pm Q(x) $ is continuous at $a$. ii. $S(x) = c \cdot P(x)$ is continuous at $a$. iii $S(x) =P(x)\cdot Q(x)$ is continuous at a$a$ . |
v. If $Q(a) \ne 0$ then $ S(x) =\frac {P(x)}{Q(x)}$ is continuous at $a$. vi. If $R(x)$ is between $P(x)$ and $Q(x)$ for the interval and $P(a)= Q(a)$ then $R$ is continuous at $a$ |