prove the statement using the epsilison, delta definition of limit:
limx® 3 x/5=3/5
how would i go about solving this?
$\displaystyle \lim_{x \to 3} \frac{x}{5} = \frac{3}{5}$
Going by definition:
$\displaystyle \forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $\displaystyle 0 < {\color{red}\mid x - 3 \mid} < \delta$ it follows that $\displaystyle \left| \frac{x}{5} - \frac{3}{5} \right| < \epsilon$
Looking at:
$\displaystyle \begin{array}{rcl}\left| \displaystyle \frac{x}{5} - \displaystyle \frac{3}{5} \right| & < & \epsilon \\ \displaystyle \frac{1}{5} {\color{red}\mid x - 3 \mid} & < & \epsilon \\ & \vdots & \end{array}$
It should be clear what you should set $\displaystyle \delta$ to be.