Question

Prove the statement using the the epsilon and delta definition of limit.

lim x/5= 3/5
x->3

Answer 1

f(x)-L < e ; |x-3|<d

Answer 2

lim x-> 3 means for all delta > 0, there exists an epsilon > 0 such that |x-3| < epsilon then |x/5-3/5| < delta

from |x/5-3/5| we see that (-1)delta < (x-3)/5 < delta
multiply through by 5 to get |(x-3)| < (5)x(delta)
and as stated earlier |x-3| < epsilon
so epsilon < 5delta

Answer 3

x/5 = 3/5 ; *5

x = 3

x-3

Answer 4

Why is it -3?

Answer 5

its just the definition as x->c

Answer 6

oh ok

Answer 7

when x-c = 0 that means you reach the spot

Answer 8

from the definition |x-a| there there is |f(x)-L| < delta

Answer 9

since we dont care about the value at x=c; as the distance between x and c gets closer and closer; we get a delta of |x-c|

Answer 10

ah ok, too many freakin laws to remember

Answer 11

:) yeah, but if you can see them for what they are, they are simpler.

spose your at x=2 and wherever I am is c=10

as i get closer to you, the distance between us is measured by |x-c|;
|2-10| = 8

i sneak closer to you and im at c=4, the distance between us is:

|x-c| = |2-4| = 2; and so the measure of delta can be between 0 and x=c or there about

Answer 12

err...

\[0<|x-c|<\delta\]