Limits of Functions
The limits of a functions determine how close the values of f(x) are to some real number L, as x approaches c, although not neccessarily at c itself.
Let f \colon D \to \mathbb{R} and let c be an accumulation point of D. We say that a real number L is a limit of f at c if for every \varepsilon > 0, there exists \delta > 0 such that
|x-c| < \delta \implies |f(x) - L| < \varepsilon
In English, a function has a limit L at c if, as x gets arbitrarily close to c, f(x) gets arbitrarily close to L.
Let f \colon D \to \mathbb{R} and c be an accumulation point of D. Then \lim_{x\to c}f(x) = L if and only if for each neighborhood of V of L there exists a deleted neighborhood U of c such that f(U \cap D) \subseteq V.
We start with the forward direction. For all \varepsilon > 0, let V = N(L,\varepsilon). Because we have \lim_{x\to c}f(x) = L, there exists some \delta such that |x-c| < \delta implies |f(x) - L| < \varepsilon, or f(x) \subseteq V. Let U = N(c,\delta)/c. The neighborhood U contains all points x \not = c such that |x-c| < \delta. Thus |f(U \cap D) - L| < \varepsilon, and f(U \cap D) \subseteq V.
Now we use the backwards direction. By letting V = N(L,\varepsilon) for all \varepsilon > 0, we have that there exists a deleted neighborhood U of c is such that f(U \cap D) \subseteq V. We let \delta represent the radius of U. For each point |x-c| < \delta, we have that f(U \cup D) \subseteq V. Thus |x -c| < \delta implies |f(x) - L| < \varepsilon.
Sequential Criterion for Limits
Limits of seqeunces and functions are closely related.
Let f \colon D \to \mathbb{R}, and c be an accumulation point of D. Then, \lim_{x\to c} f(x) = L if and only if for every sequence (s_n) in D that converges to c with s_n \not = c for all n, the sequence (f(s_n)) converges to L.
Let (s_n) be a sequence in D that converges to c and s_n \not = c for all n, and \lim_{x\to c} f(x) = L.
We start with the forward direction. We want to show that \lim_{n\to \infty} f(s_n) = L. By assumption \lim_{x\to c} f(x) = L, so given any \varepsilon > 0, there exists \delta > 0 such that 0 < |x-c| < \delta \implies |f(x) - L| < \varepsilon. Since (s_n) converges to c, there exists a natural number N such that n \geq N implies |s_n - c| < \delta. Thus for n \geq N we have 0 < |s_n - c| < \delta and s_n \in D such that |f(s_n) - L| < \varepsilon. Thus \lim_{n \to\infty} = L.
If f \colon D \to \mathbb{R} and if c is an accumulation point of D, then f can have only one limit at c.
Suppose that \lim_{x \to c}f(x) = L_1 and \lim_{x \to c} f(x) = L_2. By Theorem 2.1.2, every sequence (f(s_n)) converges to L (Granted that s_n \not = c and s_n converges to c). By Theorem 1.1.3, If (f(s_n)) converges, then it’s limit is unique. Therefore if (f(s_n)) converges to both L_1 and L_2, then L_1 = L_2.
Let f \colon D \to \mathbb{R}, and c be an accumulation point of D. Then the following are equivalent
(a) f does not have a limit at c
(b) There exists a sequence (s_n) in D with each s_n \not = c such that (s_n) converges to c, but (f(s_n)) is not convergent in \mathbb{R}.
If (a) was false (If f did have a limit at c), (b) would neccessarily be false by Theorem 2.1.2, and vice versa.
Let f(x) = \sin(1/x) for x > 0.
Claim: \lim_{x\to 0} \sin(1/x) does not exists.
We prove this by considering the sequence, s_n = \frac{2}{\pi n}. It is clear \lim s_n = 0. But the sequence (f(s_n)) is constantly oscillating between 1,0,-1,0,1,\dots, and does not converge to any real number L. Therefore by Theorem 2.1.4, \sin(1/x) does not have a limit at 0.
Let f \colon D \to \mathbb{R} and g \colon D \to \mathbb{R}. We define the sum f + g and the product fg to be the functions from D to \mathbb{R} given By
(f+g)(x) = f(x) + g(x) \quad \text{ and } \quad (fg)(x) = f(x)\cdot g(x)
for all x \in D. If k \in \mathbb{R}, then the multiple kf \colon D \to \mathbb{R} is the function defined by
(kf)(x) = k \cdot f(x), \quad \forall x \in D.
If g(x) \not = 0 for all x \in D, then the quotient f/g \colon D \to \mathbb{R} is the function defined by
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, \quad \forall x \in D.
The above ‘definitions’ are mostly just clarity on notation.
Let f \colon D \to \mathbb{R} and g \colon D \to \mathbb{R}, and let c be an accumulation point of D. If \lim_{x\ to c} f(x) = L and \lim_{x\to c}g(x) = M, and k \in \mathbb{R} then
(a) \lim_{x\to c}(f + g)(x) = L + M,
(b) \lim_{x \to c}(fg)(x) = LM,
(c) \lim_{x\to c}(kf)(x) = kL,
(d) If g(x) \not = 0 for all x \in D and M \not = 0, \lim_{x\to c}(f/g)(x) = L/M.
Tedious proof
For any polynomial P(x), we have that \lim_{x \to c} P(x) = P(c).
Apply previous theorem repeatadly
Let f \colon D \to \mathbb{R} where c is an accumulation point of D.
Suppose \lim_{x\to c} f(x) > 0. There exists a deleted neighborhood U of c, s.t. f(x) > 0, \forall x \in U \cap D.
Let L = \lim_{x\to c} f(x) > 0. Consider the neighborhood N(L,L/2). By the first theorem today, \exists a deleted neighborhood U of c such that f(x) is contained in this neighborhood for all x \in U \cap D. Therefore if L is positive, theres always a neighborhood of positive numbers.
One-Sided Limits
TO-DO
Practice
Recap
In this section we proved the following theorems and results.
Let f \colon D \to \mathbb{R} and c be an accumulation point of D. Then \lim_{x\to c}f(x) = L if and only if for each neighborhood of V of L there exists a deleted neighborhood U of c such that f(U \cap D) \subseteq V.
Let f \colon D \to \mathbb{R}, and c be an accumulation point of D. Then, \lim_{x\to c} f(x) = L if and only if for every sequence (s_n) in D that converges to c with s_n \not = c for all n, the sequence (f(s_n)) converges to L.
If f \colon D \to \mathbb{R} and if c is an accumulation point of D, then f can have only one limit at c.
Let f \colon D \to \mathbb{R}, and c be an accumulation point of D. Then the following are equivalent
(a) f does not have a limit at c
(b) There exists a sequence (s_n) in D with each s_n \not = c such that (s_n) converges to c, but (f(s_n)) is not convergent in \mathbb{R}.
Let f \colon D \to \mathbb{R} and g \colon D \to \mathbb{R}, and let c be an accumulation point of D. If \lim_{x\ to c} f(x) = L and \lim_{x\to c}g(x) = M, and k \in \mathbb{R} then
(a) \lim_{x\to c}(f + g)(x) = L + M,
(b) \lim_{x \to c}(fg)(x) = LM,
(c) \lim_{x\to c}(kf)(x) = kL,
(d) If g(x) \not = 0 for all x \in D and M \not = 0, \lim_{x\to c}(f/g)(x) = L/M.
For any polynomial P(x), we have that \lim_{x \to c} P(x) = P(c).
Let f \colon D \to \mathbb{R} where c is an accumulation point of D.
Suppose \lim_{x\to c} f(x) > 0. There exists a deleted neighborhood U of c, s.t. f(x) > 0, \forall x \in U \cap D.