# How do you prove that lim_(x->5)x^2!=24 using limit definition?

## How do you prove that ${\lim}_{x \to 5} {x}^{2} \ne 24$ using limit definition?

Dec 22, 2017

Pose $x = \xi + 5$ and consider the quantity:

$\left\mid {x}^{2} - 24 \right\mid = \left\mid {\left(\xi + 5\right)}^{2} - 24 \right\mid$

$\left\mid {x}^{2} - 24 \right\mid = \left\mid {\xi}^{2} + 10 \xi + 25 - 24 \right\mid$

$\left\mid {x}^{2} - 24 \right\mid = \left\mid {\xi}^{2} + 10 \xi + 1 \right\mid$

Using the reverse triangular inequality we can state that:

$\left(1\right) \text{ } \left\mid {x}^{2} - 24 \right\mid \ge \left\mid 1 - \left\mid {\xi}^{2} + 10 \xi \right\mid \right\mid$

Consider now any $\delta > 0$.
We have that for $x \in \left(5 - \delta , 5 + \delta\right)$ then $\left\mid \xi \right\mid < \delta$,

and based on the triangular inequality:

$\left\mid {\xi}^{2} + 10 \xi \right\mid \le {\xi}^{2} + 10 \left\mid \xi \right\mid < {\delta}^{2} + 10 \delta$

If we choose $\delta < \frac{1}{22} < 1$ then ${\delta}^{2} < \delta$ and:

$\left\mid {\xi}^{2} + 10 \xi \right\mid < 11 \delta < 11 \cdot \frac{1}{22} = \frac{1}{2}$

so that based on $\left(1\right)$:

$\left\mid {x}^{2} - 24 \right\mid \ge \left\mid 1 - \frac{1}{2} \right\mid = \frac{1}{2}$

So if we choose any $\epsilon$ such that $0 < \epsilon < \frac{1}{2}$ we can state that for $x \in \left(5 - \frac{1}{22} , 5 + \frac{1}{22}\right)$ we have that:

$\left\mid {x}^{2} - 24 \right\mid \ge \epsilon$