Prove:

$\underset{x\to 2}{lim}5 x - 1 \= 9$

Note: $c \= 2 \, L \=9\, f\left(x\right) \= 5 x-1$.

Remember you are trying to prove that:

For all $\mathrm{\ε} \> 0$, there exists a $\mathrm{\δ} \>0$such that:

if $0 < \left| x - 2\right| < \mathrm{\&delta;}$then $\left|5 x-1 -9\right| < \mathrm{\&epsilon;}$.

Step1: Determine what to choose for $\mathrm{\&delta;}\&period;$

$\left|f\&lpar;x\&rpar;-L \right| < \mathrm{\&epsilon;}$

$\left|5 x-1 -9 \right| < \mathrm{\&epsilon;}$

Substitute all values into $\left|f\&lpar;x\&rpar;-L \right| < \mathrm{\&epsilon;}$.

$\left|5 x-10\right| < \mathrm{\&epsilon;}$

$5\left| x-2\right| < \mathrm{\&epsilon;}$

$\left| x-2\right| < \frac{\mathrm{\&epsilon;}}{5}$

The relation has been simplified to the form $\left|x-c\right| < \mathrm{\&delta;}$, if you choose $\mathrm{\&delta;}\&equals;\frac{\mathrm{\&epsilon;}}{5}$.

Step 2: Assume  $\left|x\mathit{-}c\right|\mathit{ }\mathit{<}\mathit{ }\mathrm{\&delta;}$, and use that relation to prove that $\left|f\&lpar;x\&rpar;-L \right| < \mathrm{\epsilon }$.

$\left|x-c\right|<\mathrm{\&delta;}$

$\left|x-2\right|<\frac{\mathrm{\&epsilon;}}{5}$

Substitute values for $c$ and $\mathrm{\&delta;}$.

$5\left|x-2\right|<\mathrm{\&epsilon;}$

$\left|5 x-10\right| < \mathrm{\&epsilon;}$

$\left|\left(5 x-1\right)- 9\right| < \mathrm{\&epsilon;}$

$\left|f\&lpar;x\&rpar;-L \right| < \mathrm{\&epsilon;}$