# How do you prove that the limit of 5x^2 =5 as x approaches 1 using the epsilon delta proof?

Jul 27, 2018

Evaluate the difference:

$\left\mid 5 {x}^{2} - 5 \right\mid = 5 \left\mid {x}^{2} - 1 \right\mid$

$\left\mid 5 {x}^{2} - 5 \right\mid = 5 \left\mid x - 1 \right\mid \left\mid x + 1 \right\mid$

Now for any $\epsilon > 0$ choose ${\delta}_{\epsilon} < \min \left(1 , \frac{\epsilon}{15}\right)$.

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

$0 < x < 2$

and:

$1 < 1 + x < 3$

Then $\left\mid 1 + x \right\mid = \left(1 + x\right) < 3$ and so:

$\left\mid 5 {x}^{2} - 5 \right\mid = 5 \left\mid x - 1 \right\mid \left\mid x + 1 \right\mid < 15 \left\mid x - 1 \right\mid$

On the other hand, as ${\delta}_{\epsilon} < \frac{\epsilon}{15}$ then for $x \in \left(1 - {\delta}_{\epsilon} , 1 + {\delta}_{\epsilon}\right)$ we have that:

$\left\mid x - 1 \right\mid < {\delta}_{\epsilon} < \frac{\epsilon}{15}$

So:

$\left\mid 5 {x}^{2} - 5 \right\mid < 15 \left\mid x - 1 \right\mid$

$\left\mid 5 {x}^{2} - 5 \right\mid < 15 \frac{\epsilon}{15}$

$\left\mid 5 {x}^{2} - 5 \right\mid < \epsilon$

which proves the limit.