prove the statement using the epsilison, delta definition of limit:

limx® 3 x/5=3/5

how would i go about solving this?

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- Sep 14th 2008, 07:31 PMDroManstuck on this problem for cal 1
prove the statement using the epsilison, delta definition of limit:

limx® 3 x/5=3/5

how would i go about solving this? - Sep 14th 2008, 07:38 PMo_O
$\displaystyle \lim_{x \to 3} \frac{x}{5} = \frac{3}{5}$

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. - Sep 16th 2008, 05:50 AMDroMan
could you offer any tips on proving problems like this. Do you use a method? Anything that can help me grasp this concept would be greatly appreciated.

- Sep 16th 2008, 06:50 AMKrizalid
See my signature.