¡@

Limit Laws

¡@

Limit Laws

Suppose that c is a constant and limit(f(x),x = a) = L[1] and limit(g(x),x = a) = L[2] . Then

¡@

(1).(Sum Law) limit([f(x)+g(x)],x = a) = L[1]+L[2] .

(2).(Scalar Multiple Law) limit(c*f(x),x = a) = c*L[1] .

¡@

Remark :

From (1) and (2), we get limit([f(x)-g(x)],x = a) = L[1]-L[2] .

¡@

(3).(Product Law) limit([f(x)*g(x)],x = a) = L[1]*L[2] .

If we use the Product Law repeatedly with f(x) = g(x) , we obtain the following Law.

¡@

limit([f(x)]^n,x = a) = limit(f(x),x = a)^n where n is a positive integer.

(4). If L[2] <> 0 , then limit(1/g(x),x = a) = 1/L[2] .

¡@

Remarks :

1. From (3) and (4), we get

(Quotient Law) limit(f(x)/g(x),x = a) = L[1]/L[2] , if L[2] <> 0 .

¡@

2. Note that the condition L[2] <> 0 implies that g( x ) is bounded away from 0 for x near a . In fact, there exists a positive number d such that abs(L[2])/2 <= abs(g(x)) whenever x <> a and abs(x-a) < d . Hence, 1/g(x) is well-defined for x <> a near a .

¡@

[Maple Plot]

Suppose that limit(g(x),x = a) = 0 , what can we say about the behavior of g( x ) near a ?

Intuitively, limit(g(x),x = a) = 0 implies that abs(g(x)) can be as small as we want by taking x sufficiently close to a (but not equal to a ). Hence, abs(1/g(x)) can be as large as we want by taking x sufficiently close to a (but not equal to a ). We would conjecture that limit(abs(1/g(x)),x = a) = infinity . Here is the proof :

Given M > 0, since limit(g(x),x = a) = 0 , there exists 0 < delta , such that abs(g(x)) < 1/M whenever 0 < abs(x-a) < delta .We get,

¡@

M < abs(1/g(x)) whenever 0 < abs(x-a) < delta .

¡@

Hence, limit(abs(1/g(x)),x = a) = infinity .

¡@

¡@

Questions :

1. Suppose that both limit(f(x),x = a) and limit([f(x)+g(x)],x = a) exist, does limit(g(x),x = a) exist ?

2. Suppose that both limit(f(x),x = a) and limit([f(x)*g(x)],x = a) exist, does limit(g(x),x = a) exist ?

3. Suppose that limit(f(x)/g(x),x = a) exists and limit(g(x),x = a) = 0 , does limit(f(x),x = a) exist ? If so, what is the limit ?

¡@

¡@

4. Suppose that c is a constant and limit(f(x),x = a) = L[1] and limit(g(x),x = a) = infinity .

What can we say about the following limits ?

limit([f(x)+g(x)],x = a) , limit(cg(x),x = a) , limit([f(x)*g(x)],x = a) and limit(f(x)/g(x),x = a) .

¡@

¡@

5. Suppose that limit(f(x),x = a) = infinity and limit(g(x),x = a) = infinity . What can we say about the following limits ?

limit([f(x)+g(x)],x = a) , limit([f(x)-g(x)],x = a) , limit([f(x)*g(x)],x = a) and limit(f(x)/g(x),x = a) .

¡@

Theorem If f(x) <= g(x) when x is near a (except possibly at a ) and both limit(f(x),x = a) and limit(g(x),x = a) exist, then

¡@

limit(f(x),x = a) <= limit(g(x),x = a) .

¡@

Proof : Let limit(f(x),x = a) = L[1] and limit(g(x),x = a) = L[2] . Suppose that L[2] < L[1] .

¡@

¡@

[Maple Plot]

Taking epsilon = (L[1]-L[2])/2 , we have that there exists 0 < delta[1] and 0 < delta[2] such that

¡@

abs(f(x)-L[1]) < (L[1]-L[2])/2 whenever 0 < abs(x-a) < delta[1] ,

¡@

and

¡@

abs(g(x)-L[2]) < (L[1]-L[2])/2 whenever 0 < abs(x-a) < delta[2] .

Let delta = min(delta[1],delta[2]) , if 0 < abs(x-a) < delta then 0 < abs(x-a) < delta[1] and 0 < abs(x-a) < delta[2] , so we get

¡@

g(x) < (L[1]-L[2])/2 < f(x) .

¡@

But this contradicts f(x) <= g(x) . Thus, the inequality L[2] < L[1] must be false. Therefore, L[1] <= L[2] .

¡@

Question : Suppose that f(x) < g(x) when x is near a (except possibly at a ) and both limit(f(x),x = a) and limit(g(x),x = a) exist, is it true that

¡@

limit(f(x),x = a) < limit(g(x),x = a) ?

The Squeeze Theorem

If h(x) <= f(x) and f(x) <= g(x) when x is near a (except possibly at a ) and

¡@

limit(h(x),x = a) = limit(g(x),x = a) = L

¡@

then limit(f(x),x = a) = L .

¡@


[Maple Plot]

Proof : Given 0 < epsilon , since limit(h(x),x = a) = limit(g(x),x = a) = L , there exists 0 < delta[1] and 0 < delta[2] such that

¡@

abs(g(x)-L) < epsilon whenever 0 < abs(x-a) < delta[1] ,

and

¡@

abs(h(x)-L) < epsilon whenever 0 < abs(x-a) < delta[2] .

¡@

Let delta = min(delta[1],delta[2]) , if 0 < abs(x-a) < delta then 0 < abs(x-a) < delta[1] and 0 < abs(x-a) < delta[2] , so we get

¡@

L-epsilon < h(x) <= f(x) and f(x) <= g(x) < L+epsilon .

¡@

Therefore, abs(f(x)-L) < epsilon whenever 0 < abs(x-a) < delta .

¡@

Example Show that limit(x*sin(1/x),x = 0) = 0 .

¡@

¡@

[Maple Plot]

Solution :

Note that since limit(sin(1/x),x = 0) does not exist, we can not use the Product Law.

Since

¡@

abs(sin(1/x)) <= 1 for all x <> 0

¡@

we have , as illustrated above,

¡@

-abs(x) <= x*sin(1/x) and x*sin(1/x) <= abs(x) for all x <> 0 .

¡@

It is easy to show that limit(abs(x),x = 0) = 0 and limit(-abs(x),x = 0) = 0 . By the Squeeze Theorem, we get limit(x*sin(1/x),x = 0) = 0 .

¡@

Question :

1. Suppose that limit(f(x),x = a) = L , is it true that limit(abs(f(x)),x = a) = abs(L) ?

2. Suppose that limit(abs(f(x)),x = a) exists, is it true that limit(f(x),x = a) also exists ?

3. Suppose that limit(abs(f(x)),x = a) = 0 , is it true that limit(f(x),x = a) = 0 ?

¡@

Download MapleWorksheet

¡@

¡@