# Question #4c5ac

Jun 20, 2017

See below.

#### Explanation:

Prove: ${\lim}_{x \to - 3} {x}^{2} + 4 x + 1 = - 2$

Work (not part of proof):

$0 < | x + 3 | < \delta$, $| \left({x}^{2} + 4 x + 1\right) + 2 | < \epsilon$

We need to manipulate the $| \left({x}^{2} + 4 x + 1\right) + 2 | < \epsilon$ to be $| x + 3 | < \text{something}$ to set delta equal to that term:
$| \left({x}^{2} + 4 x + 1\right) + 2 | < \epsilon$

$| {x}^{2} + 4 x + 3 | < \epsilon$

$| \left(x + 3\right) \left(x + 1\right) | < \epsilon$

$| x + 3 | < \frac{\epsilon}{|} x + 1 |$

Since we cannot have an $x$ term with epsilon, we let $\delta = 1$ and solve for the value of $x + 1$:
$0 < | x + 3 | < 1$

$- 1 < x + 3 < 1$

$- 3 < x + 1 < - 1$

Here, we choose the larger value since if we chose the smaller value, the $- 3$ would not be included, so:
$| x + 1 | < 3$

Therefore,
$| x + 3 | < \frac{\epsilon}{3}$

Proof:

$\forall$ $\epsilon > 0$, $\exists$ $\delta > 0$ such that:
if $0 < | x + 3 | < \delta$, then $| \left({x}^{2} + 4 x + 1\right) + 2 | < \epsilon$.
Given $0 < | x + 3 | < \delta$, let $\epsilon = \min \left(1 , \frac{\epsilon}{3}\right)$:

$0 < | x + 3 | < \frac{\epsilon}{3}$

$0 < 3 | x + 3 | < \epsilon$

$0 < | x + 1 | | x + 3 | < \epsilon$

$0 < | {x}^{2} + 4 x + 3 | < \epsilon$

$0 < | \left({x}^{2} + 4 x + 1\right) + 2 | < \epsilon$

$\therefore$ ${\lim}_{x \to - 3} {x}^{2} + 4 x + 1 = - 2$

Jun 20, 2017

Since $f \left(x\right) = {x}^{2} + 4 x + 1$ is a polynomial, it is continuous for every $x \in \mathbb{R}$, then:

${\lim}_{x \to - 3} f \left(x\right) = {\left(- 3\right)}^{2} + 4 \cdot \left(- 3\right) + 1 = 9 - 12 + 1 = - 2$

We want to prove that given $\epsilon > 0$ we can find a corresponding ${\delta}_{\epsilon}$ such that:

$\left\mid x - \left(- 3\right) \right\mid < {\delta}_{\epsilon} \implies \left\mid {x}^{2} + 4 x + 1 - \left(- 2\right) \right\mid < \epsilon$

that is:

$\left\mid x + 3 \right\mid < {\delta}_{\epsilon} \implies \left\mid {x}^{2} + 4 x + 3 \right\mid < \epsilon$

Evaluate:

$\left\mid {x}^{2} + 4 x + 3 \right\mid = \left\mid \left(x + 3\right) \left(x + 1\right) \right\mid = \left\mid x + 3 \right\mid \left\mid x + 1 \right\mid$

Now for $\left\mid x + 3 \right\mid < {\delta}_{\epsilon}$ we have:

$\left\mid x + 1 \right\mid = \left\mid x + 3 - 2 \right\mid$

and based on the triangular inequality:

$\left\mid x + 1 \right\mid \le \left\mid x + 3 \right\mid + \left\mid 2 \right\mid$

that is:

$\left\mid x + 1 \right\mid < {\delta}_{\epsilon} + 2$

Given an arbitrary number $\epsilon > 0$ we can choose then ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{3}\right)$ and we have that because ${\delta}_{\epsilon} < 1$:

$\left\mid x + 1 \right\mid < 2 + {\delta}_{\epsilon} < 2 + 1 = 3$

and because ${\delta}_{\epsilon} < \frac{\epsilon}{3}$

$\left\mid x + 3 \right\mid < {\delta}_{\epsilon} < \frac{\epsilon}{3}$

but then:

$\left\mid x + 3 \right\mid \left\mid x + 1 \right\mid < 3 \times \frac{\epsilon}{3} = \epsilon$

and in conclusion:

$\left\mid {x}^{2} + 4 x + 3 \right\mid < \epsilon$

which proves the point.