# How do you use the epsilon delta definition to find the limit of (x^2 + 4x)  as x approaches 2?

Jul 12, 2017

Given any $\epsilon > 0$, if we choose ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{9}\right)$ then:

$x \in \left(2 - {\delta}_{\epsilon} , 2 + {\delta}_{\epsilon}\right) \implies \left\mid {x}^{2} + 4 x - 12 \right\mid < \epsilon$

so:

${\lim}_{x \to 2} {x}^{2} + 4 x = 12$

#### Explanation:

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

${\lim}_{x \to 2} f \left(x\right) = f \left(2\right) = {2}^{2} + 4 \times 2 = 4 + 8 = 12$

Evaluate now:

$\left\mid f \left(x\right) - 12 \right\mid = \left\mid {x}^{2} + 4 x - 12 \right\mid$

$\left\mid f \left(x\right) - 12 \right\mid = \left\mid \left(x - 2\right) \left(x + 6\right) \right\mid$

$\left\mid f \left(x\right) - 12 \right\mid = \left\mid x - 2 \right\mid \left\mid x + 6 \right\mid$

Given now $\epsilon > 0$, choose ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{9}\right)$

For $x \in \left(2 - {\delta}_{\epsilon} , 2 + {\delta}_{\epsilon}\right)$ we have that:

$\left\mid x - 2 \right\mid < {\delta}_{\epsilon} < \frac{\epsilon}{9}$

And as $\left(x + 6\right)$ is positive in the interval:

$\left\mid x + 6 \right\mid = x + 6 \le 2 + {\delta}_{\epsilon} + 6 < 2 + 1 + 6 = 9$

Then:

$\left\mid f \left(x\right) - 12 \right\mid = \left\mid x - 2 \right\mid \left\mid x + 6 \right\mid < \frac{\epsilon}{9} \times 9 = \epsilon$