Limits

 

We have learned that a line L is tangent to the circle at a point P if L passes through P perpendicular to the radius at P , as shown in the graph below.

 

[Maple Plot]

What does it mean to say a line L is tangent to some other curve C ? To define tangency for general curves, we need a dynamic interpretation of the tangency for a circle. The tangent line to a circle at a point P can be viewed as a result of the secant lines through P and near by points Q as Q moving toward P along the circle, as show in the animations below. (A line joining two points on the curve is a secant line to the curve.)

 

[Maple Plot]

 

 

[Maple Plot]

 

 

Suppose we want to find the equation of the tangent line to the parabola y = x^2 at the point P(1,1) . Since the tangent line passes through P(1,1) , we will be able to to get the equation of the tangent line as soon as we know its slope m . The difficulty is that we need two points to compute the slope. However, we can compute an approximation to m by choosing a nearby point Q(x,x^2) , where x <> 1 , on the parabola and compute the slope m[P*Q] of the secant line P*Q .

m[P*Q] = (x^2-1)/(x-1)

 

[Maple Plot]

Here we compute the values of m[P*Q] for several values of x close to 1 .

 

matrix([[x, m[PQ], x, m[PQ]], [1.500000000, 2.50000...

 

It appears from the table above that the closer Q is to P , the closer x is to 1 , the closer m[P*Q] is to 2 . It suggests that the slope of the tangent line should be 2 . (We shall discuss more about tangent lines in a later secession.)

 

[Maple Plot]

 

 

[Maple Plot]

 

Definition of a Limit

 

Given a function f defined for all x near a , except possibly at a . We say that

 

" the limit of f( x ), as x approaches a , equals L "

 

if we can make the values of f( x ) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a , but not equal to a . In this case, we write

 

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

 

We see from the table that the values of f(x) = (x^2-1)/(x-1) can get quite close to 2 , can we make the values of f(x) as close to 2 as we like by taking x to be sufficiently close to 1 ?

 

If we want the distance between f(x) and 2 to be less than 1/(10^5) , how close to 1 do we have to take x ? And what happens if we want the distance between f(x) and 2 to be less than 1/(10^10) ?

 

In other words, how close to 1 do we have to take x so that

 

abs((x^2-1)/(x-1)-2) < 1/(10^5) ( or 1/(10^10) ) ?

 

Notice that f(x) is not defined when x = 1 and f(x) = x+1 when x <> 1 . Hence, we have

 

abs((x^2-1)/(x-1)-2) = abs(x-1)

 

which is the distance from x to 1 . For the first case, we can simply take x so that 0 < abs(x-1) < 1/(10^5) , that is, the distance from x to 1 is less than 1/(10^5) . For the second case, we can take x so that 0 < abs(x-1) < 1/(10^10) .

[Maple Plot]

 

Similarly, if we want the distance between f(x) and 2 to be less than a given number epsilon , we can take x so that 0 < abs(x-1) < epsilon . Therefore,

 

limit((x^2-1)/(x-1),x = 1) = 2

 

From the argument above, we see that limit(f(x),x = a) does not depends whether f is defined at x = a nor how f is defined at x = a . Consequently, given any number c , if function g is defined by

 

g(x) = x+1 if x <> 1 and g(1) = 3 ,

then

limit(g(x),x = 1) = 2

 

In particular, we have

 

limit(x+1,x = 1) = 2

 

 

Example 1 Investigate limit(2*x+1,x = 2) .

 

 

 

[Maple Plot]

The graph above suggests that the values of 2*x+1 approaches 5 as approaches 2 . We guess that

 

limit(2*x+1,x = 2) = 5

 

How close to 2 do we have to take x so that abs(2*x+1-5) < 1/2 ?

Notice that abs(2*x+1-5) < 1/2 if and only if abs(x-2) < 1/4 . That means we can take so that abs(x-2) < 1/4 . Geometrically, we can also see this by the graph below.

 

 

[Maple Plot]

Since y = 2*x+1 is a line of slope 2 , we can see from the graph above that it suffices to take x so that

 

abs(x-2) < 1/4

 

Moreover, if we want the distance between 2*x+1 and 5 to be less than a given number epsilon , we can take x so that 0 < abs(x-2) < epsilon/2 . Therefore, limit(2*x+1,x = 2) = 5 .

 

 

Example 2 Investigate limit((sqrt(t^2+9)-3)/(t^2),t = 0) .

Let's take a look at the values of (sqrt(t^2+9)-3)/(t^2) for several values of t close to 0 . Since (sqrt(t^2+9)-3)/(t^2) is even, we can take only positive values of t .

 

 

matrix([[t, (sqrt(t^2+9)-3)/t^2], [.5000000000, .16...

 

As t approaches 0 , the value of the function seem to approach 0 .166666 ... and so we guess that

 

limit((sqrt(t^2+9)-3)/(t^2),t = 0) = 1/6

 

What would have happened if we had taken even smaller values of t ?

 

matrix([[t, (sqrt(t^2+9)-3)/t^2], [.1000000000e-2, ...

 

Here is a graph of y = (sqrt(t^2+9)-3)/(t^2) for t close to 0 within 0.01.

 

[Maple Plot]

What happen to the plot ?? Does this mean that the limit is 0 instead of 1/6 ????

 

Notice that (sqrt(t^2+9)-3)/(t^2) is not defined when t = 0 and (sqrt(t^2+9)-3)/(t^2) = 1/(sqrt(t^2+9)+3) when t <> 0 .

 

matrix([[t, 1/(sqrt(t^2+9)+3)], [.1000000000e-2, .1...

 

[Maple Plot]

 

 

So the limit should be 1/6 . What was wrong with previous calculations ? ----- The CAS (Maple) gave false values due to round up errors.

 

Can we make the values of f(x) as close to 1/6 as we like by taking t to be sufficiently close to 0 ? How close to 0 do we have to take t so that

abs(1/(sqrt(t^2+9)+3)-1/6) < 1/(10^5)

 

 

RealRange(Open(-600/49997*15^(1/2)),Open(600/49997*...

 

.7746431477e-4

Hence, we can take x so that abs(t) < 7/(10^5) ( or even smaller). But this seems to be hard way to confirm that

limit((sqrt(t^2+9)-3)/(t^2),t = 0) = 1/6

we shall do it in another way later.

 

Example 3 Investigate limit(abs(x)/x,x = 0) .

 

[Maple Plot]

Suppose that limit(abs(x)/x,x = 0) = L , for some number L (in this case, we say limit(abs(x)/x,x = 0) exists.). Since the value of abs(x)/x is always 1 as x approaches 0 from the right, L = 1 . But the value of abs(x)/x is always -1 as x approaches 0 from the left, so L = -1 . Here we get a contradiction. Therefore, limit(abs(x)/x,x = 0) does not exist.

 

One-sided limits

 

We say the limit of f(x) as x approaches a from the left is equal to L and write

 

limit(f(x),x = a,left) = L

 

if we can make the values of f( x ) arbitrarily close to L by taking x to be sufficiently close to a and x < a .

 

similarly, w e say the limit of f(x) as x approaches a from the right is equal to L and write

 

limit(f(x),x = a,right) = L

 

if we can make the values of f( x ) arbitrarily close to L by taking x to be sufficiently close to a and a < x .

 

With these notations, we have

limit(abs(x)/x,x = 0,left) = -1 and limit(abs(x)/x,x = 0,right) = 1

 

By comparing the definition of limits with the definitions of one-sided limits, we see that

 

limit(f(x),x = a) = L if and only if limit(f(x),x = a,left) = L and limit(f(x),x = a,right) = L

 

Example 4 Investigate limit(1/(x^2),x = 0) .

 

matrix([[x, x^2, 1/(x^2)], [.1000000000, .100000000...

 

Since g(x) = 1/(x^2) is an even function, we can just look at the values of g(x) for 0 < x . We see from the table above that as x gets close to 0 , x^2 gets close to 0 , and 1/(x^2) gets very large.

In fact, given a positive number M ( no matter how large M is), if we take abs(x) < 1/sqrt(M) then M < 1/(x^2) .That means the values of 1/(x^2) can be made as large as we like by taking x close enough to 0 .

 

Infinite Limits

 

Let f be a function defined on both sides of a , except possibly at a . Then we say the limit of f(x) is infinity as approaches a , and write

 

limit(f(x),x = a) = infinity

 

if the values of f(x) can be made as large as we like by taking x close enough to a , but not equal to a .

W e say the limit of f(x) is negative infinity as approaches a , and write

 

limit(f(x),x = a) = -infinity

 

if the values of f(x) can be made as large negative as we like by taking x close enough to a , but not equal to a . Similar definitions can be given for the one-sided infinite limits

 

limit(f(x),x = a,left) = infinity limit(f(x),x = a,left) = -infinity

 

limit(f(x),x = a,right) = infinity limit(f(x),x = a,right) = -infinity

 

 

From the previous discussion, we see that limit(1/(x^2),x = 0) = infinity . Let's take a look at the graph of g(x) = 1/(x^2) .

 

[Maple Plot]

It appears from the graph above that the graph of 1/(x^2) gets closer to the y -axis as x approaches to 0 .

 

[Maple Plot]


[Maple Plot]

 

 

Vertical Asymptotes

 

The line x = a is called a vertical asymptote of the curve y = f(x) if at least one of the following statements is true :

 

limit(f(x),x = a,left) = infinity limit(f(x),x = a,right) = infinity

 

limit(f(x),x = a,left) = -infinity limit(f(x),x = a,right) = -infinity

 

(a) limit(1/(x-1),x = 1,left) = -infinity , limit(1/(x-1),x = 1,right) = infinity .

 

 

[Maple Plot]

 

(b) limit(tan(x),x = Pi/2,left) = infinity , limit(tan(x),x = Pi/2,right) = -infinity .

 

 

[Maple Plot]

(c) limit(ln(x),x = 0,right) = -infinity .

 

[Maple Plot]

 

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