How do you use the epsilon delta definition to prove that the limit of #x/(6-x)=1# as #x->3#?

Question

1256 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T18:52:23

We can prove that

#abs (f(x) - 1) < epsilon # for #abs (x-3) < delta_epsilon#

if #delta_epsilon = (3epsilon)/(2-epsilon)#

This would not be necessary as:

#f(x) = x/(6-x)#

is continuous in #x=3#, so:

#lim_(x->3) f(x) = f(3) = 3/(6-3) = 3/3 =1#

However, as an exercise, we can see that:

#abs (f(x) - 1) = abs ( x/(6-x) -1 )= abs ( (x-6+x)/(6-x) )= abs (2x-6) / abs (6-x)#

Now consider #x in (3-delta, 3+delta)# with # delta > 0# so that:

#abs (x-3) < delta#.

We have:

#abs(6-x) = abs (3 +3-x) <= abs(3) + abs (3-x) < 3+delta#

and

#abs (2x-6) = abs (2(x-3)) = 2 abs(x-3) < 2delta#

Putting this two inequalities together:

#abs (f(x) - 1) < (2delta)/(3+delta)#

This means that given #epsilon < 0# if we choose #delta_epsilon# such that:

#epsilon = (2delta_epsilon)/(3+delta_epsilon)#

that is:

#delta_epsilon = (3epsilon)/(2-epsilon)#

we will have:

#abs (f(x) - 1) < epsilon # for #abs (x-3) < delta_epsilon#