is delta always presumed to be less than 1 in limit definition.?

Question

Limit definition in book is

F(x) is said to converge to L, as x approaches a, if and only if for every epsilon > 0 there is a delta > 0 such that

0 < |x-a| < delta implies |f(x) - L| < epsilon.

is the value of delta always presumed to be less than 1? because the book makes this assumption repeatedly when proving existence of limits in certain functions. Let me know

Answer 1

Actually, any "small real number" may be assumed for delta, but it usually makes the math easier to work with when you use delta = 1.

When you choose delta =1, you use this value to help establish constraints on other terms with x in the problem, and eventually this leads to an alternative value for delta.

Then the actual value of delta used is the smaller of the two possible choices ....
for example

lim(x^2) = 9 as x-->3

abs [ x^2 - 9] = abs[ x - 3]abs[ x + 3] < epsilon
and abs [ x - 3] < delta

Let's choose a delta other than 1 ....

assume delta = 2
abs [ x - 3 ] < 2
then - 2 < x - 3 < 2, so 1 < x < 5
thus
4 < x + 3 < 8
abs [ x + 3] < 8

now abs [ x - 3 ] abs [ x + 3 ] < 8 * abs [ x - 3 ] < epsilon
or abs [ x - 3 ] < epsilon / 8

Now the correct delta to use is the smaller value.... e.g. , if epsilon = 17 [ a bizarre choice, but certainly possible, even though it is traditionally a very small , + number, close to zero ] ,
then epsilon / 8 = 2.125 , so delta = 2 is the best choice.

If epsilon is < 16 , then delta = [ epsilon / 2 ] is the smaller value, and it is chosen...

By the way, the actual proof is to start with
delta = smaller of ( 2, epsilon / 8 ), and the fact that abs ( x - 3) < delta, .... and then show that if ....
epsilon / 8 is the smaller delta, you can get
abs ( f ( x ) - L ) < epsilon , and if delta = 2 is the smaller delta, for your given epsilon, that it also leads to the same conclusion ... abs ( f ( x ) - L ) < epsilon

Answer 2

Delta can be presumed to be smaller than any real number larger than 0. This is because the definition just asks for a delta to exist. And given any delta, if we pick a value smaller than delta as our new delta, that smaller value still satisfies the limit definition.

So more formally, if d' (d prime) is smaller than d, we have:
0 < |x-a| < d => |f(x)-L| < epsilon
(=> means implies)
and by definition
0 < |x-a| < d' < d => |f(x)-L| < epsilon
and therefore
0 < |x-a| < d' => |f(x)-L| < epsilon
since values between 0 and d' are a subset of the values between 0 and d.

Answer 3

It isn't always presumed. But suppose there is a delta greater than 1 such that 0 < |x-a| < delta implies |f(x) - L| < epsilon. Then obviously a smaller delta will also work. Starting with the assumption that delta is less than one may make a proof easier though.