4.1 Definition of Limit for Functions 181Example 4.1.3 Prove that lim(4x - 5) = 3.
X->2
(This example should remind you of Example 2.1.6.)
Proof. First, let f(x) = 4x - 5. Note that 2 is an interior point of V(f).
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Let c > 0. Choose 6 =
4(as suggested by Part (c) of Example 4.1.2). Then0 < Ix - 21 < 6 =? Ix - 21 < 6
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=} Ix - 21 < -
4
=? 4lx - 21 < c
=? l4x - 81 < c
=? I ( 4x - 5) - 3 I < c.Therefore, by Definition 4.1.1, lim(4x - 5) = 3. D
X->2
The next example demonstrates the role of "O < Ix - xol" in the definition
of lim f(x ) = L.
x-..xo
... 2x^2 - 18
Example 4.1.4 Use Defimt10n 4.1.l to prove that hm
3
= 12.
x->3 X -. 2x^2 - 18..
Proof. Consider f(x) =. Let E: > 0. This time 3 ¢. V(f). Note
x-3
2(x-3)(x+3) E:
that when x "I 3, f(x) = = 2x + 6. Choose 6 = -
2
. Then
x-3
0 < Ix - 31 < 6 =? x "I 3 and Ix - 3\ < 6
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=? f(x) =2x+6 and \x-3\ < 2
=? \f(x ) - 121 = l(2x + 6) - 12\ = \2x - 6\
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=? \f(x ) - 12 \ = 2\x - 31 < 2 · 2 = E:
=? \f(x ) - 121 < E:.2x^2 - 18
Therefore, lim = 12. D
X->3 X - 3Example 4.1.5 Consider the limit statement: X->2 lim x^2 = 4.(a) Find a value of 6 > 0 that will guarantee that whenever x is within
distance 6 from 2 (but not equal to 2) x^2 is within distance .01 from 4.(b) For arbitrary E: > 0, find a value of 6 > 0 that will guarantee that when-
ever x is within a distance 6 from 2 (but not equal to 2) x^2 is within a
distance c from 4.