stuck on this problem for cal 1

• Sep 14th 2008, 08:31 PM
DroMan
stuck 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, 08:38 PM
o_O
$\lim_{x \to 3} \frac{x}{5} = \frac{3}{5}$

Going by definition:
$\forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $0 < {\color{red}\mid x - 3 \mid} < \delta$ it follows that $\left| \frac{x}{5} - \frac{3}{5} \right| < \epsilon$

Looking at:
$\begin{array}{rcl}\left| \displaystyle \frac{x}{5} - \displaystyle \frac{3}{5} \right| & < & \epsilon \\ \displaystyle \frac{1}{5} {\color{red}\mid x - 3 \mid} & < & \epsilon \\ & \vdots & \end{array}$

It should be clear what you should set $\delta$ to be.
• Sep 16th 2008, 06:50 AM
DroMan
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, 07:50 AM
Krizalid
See my signature.