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 -δ and +δ, then f(x) - L lies between -ε 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 -δ and +δ, 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 -ε and +ε

(2) Since the limit of f(x) = L, we know that there exists δ

_{1}such that:

if x - a is between -δ

_{1}and +δ

_{1}, then f(x) - L is between -ε and +ε

Since by definition, if x - a is between -δ

_{1}and +δ

_{1}, then f(x) is between -L and +L.

(3) We also know that there exists δ

_{2}such that:

if x - a is between -δ

_{2}and +δ

_{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 δ

_{3}such that:

if x - a is between -δ

_{3}and +δ

_{3}, then g(x) - M is between -ε/(2L) and ε/(2L)

(5) Let δ = min(δ

_{1},δ

_{2},δ

_{3})

(6) Now, if x - a is between -δ and +δ, 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 -ε and +ε

(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 δ

_{1}and δ

_{2}such that:

if x-a lies between -δ

_{1}and +δ

_{1}, then f(x)-L lies between -ε and +ε

if x-a lies between -δ

_{2}and +δ

_{2}, then h(x)-L lies between -ε and +ε

(3) Let δ = min(δ

_{1},δ

_{2})

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

(5) If x-a in between -δ and +δ, 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 -δ and +δ, then g(x)-L is between -ε and +ε

(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 δ

_{1}such that:

if (y - L) is between -δ

_{1}and +δ

_{1}, then f(y) - f(L) is between -ε and +ε

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

_{2}such that:

if (x - a) is between - δ

_{2}and +δ

_{2}, then g(x) - L is between -δ

_{1}and +δ

_{1}

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

if (x -a ) is between -δ

_{2}and +δ

_{2}, then y - L is between -δ

_{1}and +δ

_{1}and f(g(x)) - f(L) is between -ε and +ε

(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/a

^{2}* abs(x-a)

(8) Let δ be the minimum of a/2 and a

^{2}ε/2

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

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

^{2})(a

^{2}ε/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/a

^{2}*abs(x-a)

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

^{2}ε/2

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

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

^{2})(a

^{2}ε/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

Corollary 8.1: Addition 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.

(2) Since the limit of f(x) = L, we know that there exists δ

_{1}such that:

if x - a is between -δ

_{1}and +δ

_{1}, then f(x) - L is between -ε/2 and +ε/2

Since by definition, if x - a is between -δ

_{1}and +δ

_{1}, then f(x) is between -L and +L.

(3) Since the limit of g(x) = L, we know that there exists δ

_{2}such that:

if x - a is between -δ

_{2}and +δ

_{2}, then g(x) - M is between -ε/2 and +ε/2

Since by definition, if x - a is between -δ

_{2}and +δ

_{2}, then g(x) is between -M and +M.

(4) Let δ = min(δ

_{1},δ

_{2}

_{})

(5) Now, if x - a is between -δ and +δ, 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

- Mathematical Limit, Wikipedia
- Edwards & Penny, Calculus and Analytic Geometry

## 5 comments :

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

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

thanx this really helped.

and I'm not so convinced about your quotient and reciprocal law.

Hi Kiberu,

Which step are you unclear about?

-Larry

Post a Comment