When demonstrating that the limit
as x approaches negative one of two x plus five equal three
with epsilon equal to zero point five,
which of the following delta values suffice?
This problem deals with the formal definition of a limit.
Let's begin by viewing this deifinition.
The definition of the limit as approaches a
of f of x equals L is that for every given number
epsilon greater than zero, there is a number delta
greater than zero, such that if x is
within delta units of a and x doesn't equal a,
which means x is in this interval here from a plus delta
to a minus delta, where again x doesn't equal a,
then f of x is within epsilon units of L,
meaning f of x would be in this interval here
from L minus epsilon to L plus epsilon.
We can also state this definition using inequalities
as we see here.
Remember absolute value represents a distance.
So we have the absolute value of f of x minus L,
meaning the distance between f of x and L is less
than epsilon units whenever the absolute value of x
minus a, the distance between x and a the value
we're approaching is less than delta units
and greater than zero units.
So for our problem, we're given epsilon.
We need to find a relationship between epsilon and delta.
So applying our definition, the absolute value of f of x
minus L less that epsilon would give us the absolute value
of the quantity two x plus five
minus L, which is three, is less than epsilon,
which is zero point five
whenever the absolute value of x
minus a would be the absolute value of x
minus negative one is less than delta and greater than zero.
We want to use these absolute value inequalities
to find the relationship between epsilon and delta.
So simplifying we have the absolute value of two x,
this would be plus two, less than zero point five.
Over here on the right we'd have the absolute value of x
plus one, is less than delta and greater than zero.
Now going back over here on the left,
notice how if we factor out two,
we'd have two times the absolute value of x plus one
is less than zero point five.
And then if we divide both sides by two,
we have the absolute value of x plus one
is less than zero point two five.
This is what we need in order to find our value for delta.
If the absolute value of x plus one
is less than zero point two five,
we'd have the absolute value of x plus one
is less than delta, which is equal to zero point two five.
This is also greater than zero.
So this is the largest value of delta
that would satisfy our definition
with epsilon equal to zero point five,
but of course delta could be smaller.
So really delta can be less than
or equal to zero point two five and greater than zero.
So going back to our problem,
the delta values that will work again are the delta values
that are less than or equal to zero point two five
and greater than zero.
So the only value that doesn't work would be
this first value of zero point five
because zero point five is greater than zero point two five.
This value, this value, and this value delta
are all less than equal to zero point two five
and greater than zero.
If we take a look at this graphically,
if we look at this graphically,
when we were given this limit
and that epsilon is equal to zero point five,
we wanted our function values to be
within zero point five units of the limit value L.
In order for this to occur,
we discovered that the x values had to be
within zero point two five units of negative one.
I hope you found this helpful