How do you prove that the limit of (2x^2 + 1) = 3  as x approaches 1 using the epsilon delta proof?

1 Answer
Jun 7, 2018

Let $x = 1 + \xi$ so that:

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

Given any number $\epsilon > 0$ choose ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{6}\right)$.

Then as ${\delta}_{\epsilon} < 1$ we have that $x \in \left(1 - {\delta}_{\epsilon} , 1 + {\delta}_{\epsilon}\right)$ implies that $\left\mid \xi \right\mid < 1$ and:

$\left\mid 2 + \xi \right\mid \le 2 + \left\mid \xi \right\mid < 3$

Furthermore, as ${\delta}_{\epsilon} < \frac{\epsilon}{6}$ we have that $x \in \left(1 - {\delta}_{\epsilon} , 1 + {\delta}_{\epsilon}\right)$ implies $\left\mid \xi \right\mid < \frac{\epsilon}{6}$ and then:

$\left\mid 2 {x}^{2} + 1 - 3 \right\mid = 2 \left\mid \xi \right\mid \left\mid 2 + \xi \right\mid < 2 \cdot \frac{\epsilon}{6} \cdot 3 = \epsilon$

Thus given $\epsilon > 0$ and ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{6}\right)$:

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

which proves that:

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