How do you find the Limit of #2x+5# as #x->-3# and then use the epsilon delta definition to prove that the limit is L?

Question

1419 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T19:29:16

If #x# is very close to #-3#,

then #2x# is very close to #2(-3) = -6#

and #2x+5# is very close to #-6+5 = -1#.

So #lim_(xrarr-3)(2x+5) = -1#

Proof:

Given a positive number, #epsilon#, let #delta = epsi/2#.

Then for every #x# with #0 < abs (x-(-3)) < delta#, we have #abs(x+3) < epsi/2#.

Therefore #abs((2x+5)-(-1))# which is equal to #abs(2x+6)# is equal to #2abs(x+3)# is less than #2(delta)# which is #2(epsi/2)# or just #epsi#.

Writing it in mathematics we write:

For #0 < abs (x-(-3)) < delta#, we have

#abs((2x+5)-(-1)) = abs(2x+6)#

# = 2abs(x+3)#

# < 2(delta)#

# = 2(epsi/2) = epsi#.

So, by the definition of limit, #lim_(xrarr-3)(2x+5) = -1#