Delta-Epsilon Definition of a Limit...5

[nycmathguy]

Calculus
Section 1.7

How about 24? Please, show every step along the way.

 

[MathLover1]

24.

Theorem: For any constant c and any real number a,

.

Proof:

Let ϵ > 0 be given, now we need to choose a δ > 0 such that for any x, that satisfies |x − a| < δ, the condition
|f(x) − c| < ϵ is satisfied.

Well this one is an easy one to prove because f(x) = c for any x.

So |f(x) − c| = |c − c| = 0.

So there is no game here at all, because for any given ϵ > 0, it doesn’t matter what δ I choose, for any x, in any (a−δ, a+δ) interval corresponding |f(x) − c| = |c − c| = 0 < ϵ.
So without a difficulties in this case is proved that

.

 

[nycmathguy]

24.

Theorem: For any constant c and any real number a, View attachment 2721.

Proof:

Let ϵ > 0 be given, now we need to choose a δ > 0 such that for any x, that satisfies |x − a| < δ, the condition
|f(x) − c| < ϵ is satisfied.

Well this one is an easy one to prove because f(x) = c for any x.

So |f(x) − c| = |c − c| = 0.

So there is no game here at all, because for any given ϵ > 0, it doesn’t matter what δ I choose, for any x, in any (a−δ, a+δ) interval corresponding |f(x) − c| = |c − c| = 0 < ϵ.
So without a difficulties in this case is proved that View attachment 2722.

This is simply amazing work. Great study notes.