How do you prove that the limit #(3x + 5) = -1# as x approaches -1 using the formal definition of a limit?

Answer 1

We cannot prove it. It is not true. We can prove that #lim_(xrarr-2)(3x+5)=-1#

Preliminary analysis

#lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)# if and only if
for every #epsilon > 0#, there is a #delta > 0# such that: for all #x#, #" "# if #0 < abs(x-color(green)(a)) < delta#, then #abs(color(red)(f(x))-color(blue)(L)) < epsilon#.
So we want to make #abs(underbrace(color(red)((3x+5)))_(color(red)(f(x)) )-underbrace((color(blue)(-1)))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace((color(green)(-2)))_color(green)(a))# which is equal to #abs(x+2)#.

Look at the thing we want to make small. We want to see the thing we control in the thing we want small..

#abs((3x+5)-(-1)) = abs(3x+6) = abs (3(x+2)) =abs(3)abs(x+2) =3abs(x+2)#
We want to make #3abs(x+2)# small and we control the size of #abs(x+2)#.
We can make #3abs(x+2) < epsilon# by making #abs(x+2) < epsilon/3#.
So we will choose #delta = epsilon/3#. (Any lesser #delta# would also work.)
(Detail: if #abs(x+2) < epsilon/3#, then we can multiply on both sides by the positive number #3# to get #3abs(x+2) < epsilon#.)

Now we need to actually write up the proof:

Proof

Given #epsilon > 0#, choose #delta = epsilon/3#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-(-2)) < delta#, we have
#abs(x + 2) < delta#, and
#abs((3x+5) - (-1)) = abs(3x+6)) = abs(3)abs(x+2) = 3abs(x+2) < 3delta#
[Detail: if #abs(x+2) < delta#, we can conclude that #3abs(x+2) < 2delta#. #" "# We usually do not mention this, but leave it to the reader. See below.]
And #3 delta = 3 epsilon/3 = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-(-2)) < delta#, we have #abs((3x+5) - (-1)) < epsilon#
So, by the definition of limit, #lim_(xrarr-2)(3x+5) = -1#.

We can condense a bit

Given #epsilon > 0#, choose #delta = epsilon/3#. #" "# (note that #delta# is also positive).
for every #x# with #0 < abs(x-(-2)) < delta#, we have
#abs((3x+5)-1) = abs(3x+6)#
# = abs(3(x-2))#
# = 3abs(x-2)#
# < 3delta = 3 epsilon/3 = epsilon#.
So, #abs((3x+5)-(-1)) < epsilon#.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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