1 INTRODUCTION
1.1 LIMITS
If f (x) tends to L as x tends to a, we say that the limit of f (x), as x tends
to a is L. This is usually abbreviated as;
lim f (x) = L
x→a
lim f (x) is the Value that f (x) approaches as x approaches a
x→a
Definition 1: lim f (x) = L if ∀ > 0, there is δ > 0 such that
x→a
0 < |x − a| < δ =⇒ |f (x) − L| < .
Definition 2: We say lim f (x) = ∞ if given M , there is δ > 0,
x→a
∃ 0 < |x − a| < δ =⇒ f (x) > M .
lim+ f (x) is the limit of f as x tends to a through values greater than
x→a
a (Right hand limit)
lim− f (x) is the limit of f as x tends to a through values less than a
x→a
(Left hand limit)
Theorem: Suppose that lim f (x) = L1 and lim g(x) = L2 , then;
x→a x→a
(1) lim f (x) + g(x) = L1 + L2
x→a
(2) lim λf (x) = λL1 (λ is a constant)
x→a
(3) lim f (x)g(x) = L1 L2
x→a
f (x) L1
(4) lim = provided that L2 6= 0
x→a g(x) L2
2020
, 2
Examples
(1)
lim (3x3 − x2 + 2x + 5) = 3 lim x3 − lim x2 + 2 lim x + lim 5
x→2 x→2 x→2 x→2 x→2
3 2
= 3(2 ) − 2 + 2(2) + 5
= 29
(2)
x2 − 1 0
lim =
x→1 x + 1 2
= 0
(x − 1)(x + 1)
= lim
x→1 (x + 1)
= lim (x − 1)
x→1
= 0
(3)
x2 − 1 (x − 1)(x + 1)
lim = lim
x→1 x − 1 x→1 (x − 1)
= lim (x + 1)
x→1
= 2
2020
1.1 LIMITS
If f (x) tends to L as x tends to a, we say that the limit of f (x), as x tends
to a is L. This is usually abbreviated as;
lim f (x) = L
x→a
lim f (x) is the Value that f (x) approaches as x approaches a
x→a
Definition 1: lim f (x) = L if ∀ > 0, there is δ > 0 such that
x→a
0 < |x − a| < δ =⇒ |f (x) − L| < .
Definition 2: We say lim f (x) = ∞ if given M , there is δ > 0,
x→a
∃ 0 < |x − a| < δ =⇒ f (x) > M .
lim+ f (x) is the limit of f as x tends to a through values greater than
x→a
a (Right hand limit)
lim− f (x) is the limit of f as x tends to a through values less than a
x→a
(Left hand limit)
Theorem: Suppose that lim f (x) = L1 and lim g(x) = L2 , then;
x→a x→a
(1) lim f (x) + g(x) = L1 + L2
x→a
(2) lim λf (x) = λL1 (λ is a constant)
x→a
(3) lim f (x)g(x) = L1 L2
x→a
f (x) L1
(4) lim = provided that L2 6= 0
x→a g(x) L2
2020
, 2
Examples
(1)
lim (3x3 − x2 + 2x + 5) = 3 lim x3 − lim x2 + 2 lim x + lim 5
x→2 x→2 x→2 x→2 x→2
3 2
= 3(2 ) − 2 + 2(2) + 5
= 29
(2)
x2 − 1 0
lim =
x→1 x + 1 2
= 0
(x − 1)(x + 1)
= lim
x→1 (x + 1)
= lim (x − 1)
x→1
= 0
(3)
x2 − 1 (x − 1)(x + 1)
lim = lim
x→1 x − 1 x→1 (x − 1)
= lim (x + 1)
x→1
= 2
2020