1549901369-Elements_of_Real_Analysis__Denlinger_

4.1 Definition of Limit for Functions 183

Example 4.1.6 Prove that lim x^2 = 4.

X->2

Proof. Let c. > 0. Choose o = min{l, g}. Then

E.

Ix - 21 < o ::::} Ix - 21 < 1 and Ix - 21 < -

5 E.

::::} -1 < x - 2 < 1 and Ix - 21 < -

E. 5

::::} 3 < x + 2 < 5 and Ix - 21 < -

E. 5

::::} Ix+ 21 < 5 and Ix - 21 < -

E. 5

::::} Ix+ 21 Ix - 21 < 5 · S

::::} 1x2 - 41 < c..

Therefore, lim x^2 = 4. 0

x->2

SUMMARY: HOW TO PROVE lim f(x) = L

X--+Xo

1. Let c. > 0.


2. Find a value of o > 0 that will guarantee that whenever x is
   within a distance o from xo (but not equal to x 0 ), f(x) is within
   a distance c. from L. [This is what we did in Part ( c) of Example
   4.1.2 and Part (b) of Example 4.1.5 above.]


3. Let o = the value found in Step 2.


4. Prove that for this value of o,
   't/x E D(f), 0 < Ix - xol < o::::} lf(x) - LI < c..
   [This is what we did in Examples 4.1.3, 4.1.4, and 4.1.6 above.]

Note: Step 2, although of critical importance in finding o, is not considered
part of the proof of lim f ( x) = L. It is never included when the proof is written
x --+xo
up. It may be discarded once Step 4 is completed. In fact, step 4 is usually done
by working Step 2 backwards, as demonstrated in Examples 4.1.3, 4.1.4, and
4.1.6.

Lemma 4.1.7 (Negation of lim f(x) = L): f does not have limit L
ro-+m 0
at xo if and only if

I :ic. > 0 3 Vo> 0, 3 x E D(f) 3 0 < Ix -xol < o and lf(x) - LI;::: c.. I