# How do you prove a limit of (x^2+3) as x approaches 1 equal 4? Thanks

## Please show the process using Epsilon-Delta proof. Thanks again

Jun 10, 2018

Evaluate the difference:

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

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

Let now $x = 1 + \xi$, then:

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

Given any number $\epsilon > 0$ choose:

${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{3}\right)$

For $x \in \left(1 - {\delta}_{\epsilon} , 1 + {\delta}_{\epsilon}\right)$ we have that:

$\left\mid \xi \right\mid < {\delta}_{\epsilon}$

Now, as ${\delta}_{\epsilon} < 1$ we have that:

$\left\mid 2 + \xi \right\mid \le 2 + \left\mid \xi \right\mid < 2 + 1 = 3$

and then:

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

and because ${\delta}_{\epsilon} < \frac{\epsilon}{3}$ we have that:

$\left\mid {x}^{2} + 3 - 4 \right\mid < 3 \left\mid \xi \right\mid < 3 \cdot \frac{\epsilon}{3} = \epsilon$

We can conclude that for every $\epsilon$ if we choose ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{3}\right)$ we have that:

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