lim
n→∞
1
=0
n
Proof:
For any ε > 0, we need to find N ∈ N such that for all n ≥ N , we have
1
−0 <ε
n
Since
1
n
> 0 for all n ∈ N, the condition simplifies to
1
1
< ε =⇒ n >
n
ε
Let N = ⌈ 1ε ⌉ + 1. Then, for all n ≥ N ,
1
1
1
≤
< 1 =ε
n
N
ε
Thus, by the ε-N definition, we have shown that
lim
n→∞
1
=0
n
(a) Fundamental Trig Limit: limx→0
sin(x)
x
=1
(b) Fundamental Log Limit: limx→0+ x ln(x) = 0
(c) Intermediate Value Theorem: If f is a continuous function on the
closed interval [a, b], then for any y between f (a) and f (b), there exists c
in (a, b) such that f (c) = y.
(d) Extreme Value Theorem: If f is a continuous function on a closed
interval [a, b], then f attains both a maximum and a minimum value on
[a, b].
(e) Mean Value Theorem: If f is a continuous function on the closed
interval [a, b] and differentiable on the open interval (a, b), then there exists
c in (a, b) such that
f (b) − f (a)
f ′ (c) =
b−a
3.
f (x)
cx , α ≥ 0, c ∈ R
sin(x)
cos(kx), k ̸= 0
eγx , γ ∈ R
ln(x), for x > 0
α
1
f ′ (x)
αcxα−1
cos(x)
−k sin(kx)
γeγx
1
x
4. plot of f (x) =
x3 +x2/3
x2 −1
Figure 1: Graph of f (x) =
2
x3 +x2/3
x2 −1