## Wednesday, March 08, 2006

### Mathematical Limits

One of the most important ideas in calculus is the concept of the mathematical limit. Limits relate to continuous functions and the basic idea is that if as a value the argument x approaches the value a, the difference between the limit L and f(x) can be arbitrarily small.

Definition 1: Mathematical Limit:

A function f(x) has a limit of L at point a if given any number ε, there exists a positive number δ such that:
if x-a lies between and , then f(x) - L lies between and ε

This definition is very similar to the definition of a continuous function (see here) and it is not surprising that the two concepts are very closely related.

In today's blog, I will need two definitions in order to prove the Squeeze Law relating to mathematical limits.

Definition 2: Open Interval :

x is an element of an open interval (α, β) if x is greater than α and x is less than β

Definition 3: Deleted Neighborhood

A deleted neighborhood is a set of points that result from deleting a single point in an open interval.

Lemma 1: Constant Law for Limits

if f(x) = C, then lim (x → a) f(x) = C

Proof:

(1) Let δ = 1

(2) if x - a lies between and , we know that f(x) = C.

(3) So, we know that f(x) - C = C - C = 0 which is less than any positive value ε

QED

Lemma 2: Product Law

if lim (x → a) f(x) = L and lim(x → a)g(x) = M, then
lim(x → a)[f(x)*g(x)] = L * M

Proof:

(1) Let ε be any nonzero value. We will prove that f(x)g(x) - LM lies between and

(2) Since the limit of f(x) = L, we know that there exists δ1 such that:
if x - a is between 1 and 1, then f(x) - L is between and

Since by definition, if x - a is between 1 and 1, then f(x) is between -L and +L.

(3) We also know that there exists δ2 such that:
if x - a is between 2 and 2, then f(x) - L is between -ε/(2M) and ε/(2M).

The definition for limits is that for any given positive value (ε), we can find a positive value (δ) to get the result (see above if review is needed).

(4) And there exists δ3 such that:
if x - a is between 3 and 3, then g(x) - M is between -ε/(2L) and ε/(2L)

(5) Let δ = min(δ123)

(6) Now, if x - a is between and , then:

(a) f(x) - L is between -ε/(2M) and +ε/(2M)

(b) g(x) - M is between - ε/(2L) and + ε/(2L)

(c) M[f(x) - L] is between (M)[-ε/(2M)] and (M)[+ε/(2M)] which is between -ε/2 and ε/2.

(d) f(x)[g(x) - M] is between (L)[-ε/(2L)] and (L)[+ε/(2L)] which is between -ε/2 and ε/2.

(e) If we add (c) + (d), we get:
f(x)M - LM + f(x)g(x) - f(x)M = f(x)g(x) - LM

(f) So, f(x)g(x) - LM is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2) which means that it is between and

(7) So LM is the limit for f(x)g(x).

QED

Lemma 3: Squeeze Law

Suppose f(x), g(x), h(x) are functions such that
(a) f(x) ≤ g(x) ≤ h(x) for a deleted neighborhood (α, β) where point a is removed.
(b) lim (x→ a) f(x) = L = lim(x→a)h(x).

Then:
lim (x→ a) g(x) = L

Proof:

(1) Let ε be an arbitary number.

(2) Using the definition of limits, we know that there exists δ1 and δ2 such that:

if x-a lies between 1 and 1, then f(x)-L lies between and

if x-a lies between 2 and 2, then h(x)-L lies between and

(3) Let δ = min(δ12)

(4) We know that δ is greater than 0. [By the definition of mathematical limit]

(5) If x-a in between and , we know that f(x) and h(x) are both points of the open interval (L-ε, L+ε) [Again, from the definition of mathematical limit]

(6) So L-ε is less than f(x) ≤ g(x) ≤ h(x) which is less than L + ε

(7) Combining (#5) and (#6), this gives us that for any given ε, there exists a δ such that:
if x-a is in between and , then g(x)-L is between and

(8) From (#7), L is also a limit for g(x) as x approaches a.

QED

Lemma 4: Substitution Law

If lim (x → a) g(x) = L and lim (x → L) f(x) = f(L), then lim (x → a) f(g(x)) = f(L)

Proof:

(1) Let ε be any positive real value.

(2) Because lim (y → L) f(y) = f(L), we also know that there exists a value δ1 such that:

if (y - L) is between 1 and 1, then f(y) - f(L) is between and

(3) Because lim (x → a) g(x) = L, we know that there exists a value δ2 such that:

if (x - a) is between - δ2 and 2, then g(x) - L is between 1 and 1

(4) But this means if y = g(x), then:

if (x -a ) is between 2 and 2, then y - L is between 1 and 1 and f(g(x)) - f(L) is between and

(5) This then proves that:

lim (x → a) f(g(x)) = f(L)

QED

Lemma 5: lim (x → a) (1/x) = 1/a if a ≠ 0

Proof:

(1) Let ε be any positive real number.

(2) Assume that a is greater than 0.

(3) abs(1/x - 1/a) = abs([a - x]/ax) = abs([x -a]/ax) = (1/a)abs(x-a)/abs(x)

(4) Let us assume that abs(a-x) is less than a/2.

We can do this since abs(x-a) approaches 0 as x moves toward a.

(5) Then x -a is between -a/2 and +a/2 which means that x is between a/2 and 3a/2.

(6) This gives us that abs(x) is greater than a/2 and 1/abs(x) is less than 2/a.

(7) So that abs(1/x - 1/a) = abs(x-a)*(1/a)*abs(1/x) which is less than abs(x-a)*(1/a)(2/a) = 2/a2 * abs(x-a)

(8) Let δ be the minimum of a/2 and a2ε/2

(9) Then if x - a is between and , then:

abs(1/x - 1/a) is less than (2/a2)(a2ε/2) = ε

This then proves that lim (x → a) (1/x) = 1/a for when a is greater than 0.

(10) Assume that a is less than 0

(11) Then

abs(1/x - 1/a) = abs(x-a)/(-a)*1/abs(x)

(12) If we assume that abs(x-a) is less than -a/2, then:

x - a is between -a/2 and +a/2, then x is between -3a/2 and -a/2.

(13) So abs(x) is greater than -a/2.

(14) So 1/abs(x) is less than -2/a.

(15) In this case, then:

abs(1/x - 1/a) = abs(x-a)/(-a)*1/abs(x) which is less than 1/(-a)*(-2/a)*abs(x-a) = 2/a2*abs(x-a)

(16) Let δ be the minimum of -a/2 and a2ε/2

(17) Then if x - a is between and , then:

abs(1/x - 1/a) is less than (2/a2)(a2ε/2) = ε

This then proves that lim (x → a) (1/x) = 1/a for when a is less than 0.

QED

Lemma 6: Reciprocal Law

if lim(x → a) g(x) = L and L ≠ 0, then lim (x → a) 1/g(x) = 1/L

Proof:

(1) Let f(x) = 1/x

(2) lim (x → a) f(x) = lim (x → a) (1/x)

(3) Using Lemma 5 above, we have:

lim (x → a) f(x) = 1/L = f(L)

(4) Applying the Substitution Law (Lemma 4 above) gives us:

lim (x → a) 1/g(x) = lim (x → a) f(g(x)) = f(L) = 1/L

QED

Lemma 7: Quotient Law

if lim (x → a) f(x) = L and lim (x → a) g(x) = M ≠ 0, then:

lim (x → a) f(x)/g(x) = L/M

Proof:

(1) Using the Product Law above, we have:

lim (x → a) f(x)/g(x) = lim (x → a)f(x) * lim(x → a)1/g(x)

(2) Using the Reciprocal Law above:

lim (x → a) 1/g(x) = 1/M

(3) Combining step #1 and step #2 gives us:

lim (x → a) f(x)/g(x) = L*(1/M) = L/M

QED

Lemma 8: abs(a + b - c - d)) ≤ abs(a - c) + abs(b - d)

Proof:

(1) If (a-c),(b-d) are the same sign, then abs(a + b - c -d) = abs(a -c) + abs(b - d)

(2) If (a-c),(b-d) are not the same sign, then abs(a - c + b - d) is less than abs(a -c) + abs(b-d).

QED

if lim (x → a) f(x) = L and lim(x → a)g(x) = M, then
lim(x → a)[f(x)+g(x)] = L + M

Proof:

(1) Let ε be any nonzero value.

(2) Since the limit of f(x) = L, we know that there exists δ1 such that:
if x - a is between 1 and 1, then f(x) - L is between -ε/2 and +ε/2

Since by definition, if x - a is between 1 and 1, then f(x) is between -L and +L.

(3) Since the limit of g(x) = L, we know that there exists δ2 such that:
if x - a is between 2 and 2, then g(x) - M is between -ε/2 and +ε/2

Since by definition, if x - a is between 2 and 2, then g(x) is between -M and +M.

(4) Let δ = min(δ12)

(5) Now, if x - a is between and , then:

(a) f(x) - L is between -ε/2 and +ε/2

(b) g(x) - M is between - ε/2 and + ε/2

(c) By Lemma 8 above, abs([f(x) + g(x)] - (L + M)) ≤ abs(f(x) - L) + abs(g(x) - M)

(d) abs(f(x) - L) + abs(g(x) - M) ≤ η/2 + η/2 = η

(7) So L+M is the limit for f(x)+g(x).

QED

References

Anonymous said...

You must be smart, but I feel like you just wrote all that in chinese. Didn't help me understand much.

Larry Freeman said...

It takes some time getting used to formal language if you are interested.

If it's not for you, I suggest checking out other web sites such as Wikipedia.org, cut-the-knot.org, and betterexplained.com.

-Larry

kiberu emma said...

thanx this really helped.

kiberu emma said...

Larry Freeman said...

Hi Kiberu,

Which step are you unclear about?

-Larry

TechOn top said...

I am really thankful to you for such nice information.

Max Dee said...

You really explained it well. Hats off to you. Anybody who didn't understand it can also use online limit solver to solve their mathematics problems.