Topic 1 Function and Limits

1.1 Tangent and Velocity

Exercise 1.1 Find the tangent line to the curve defined by y=1x at the point (1,1) using approximations by secant line.

Solution.

The slope of the secant line passing through (1,1) and another point (x,1x) is given by m(x)=1x1x1=1x.

When x approaches 1 the slope m(x) approaches 1 which is the slope of the tangent line.

Using the point-slope formula y=m(xx0)+y0 of the line passing thought (x0,y0) with the slope m, we can write an equation for the tangent line as follows y=(x1)+1=x+2.

Exercise 1.2 If a ball is thrown into the air with a velocity of 50 feet/second, its height in feet t seconds later is given by y=50t16t2. Estimate the instantaneous velocity when t=2 using approximations by average velocity.

Solution.

The average velocity between 2 seconds and t seconds after the ball being thrown is given by v(t)=(50t16t2)(5021622)t2=5016(t+2).

When t approaches 2 the average velocity v(t) approaches the instantaneous velocity which is v=5016(2+2)=14 feet/second.

1.2 Limits of a Function

Note: In this course, the limit limxcf(x) of a function f is defined/understood independent on how (or whether) f is defined at c. This is the most popular definition. See the Wiki page Deleted versus non-deleted limits for more detail.

Exercise 1.3 Evaluate the limit limx1g(x) using the graph of the function y=g(x) shown in the figure.

The graph of a function with a point replaced by another point

Solution.

When x goes to 1, the point (x,g(x)) on the graph goes to the empty dot whose y-coordinate is 2. So the limit limx1g(x)=2

Exercise 1.4 Evaluate the limit limx1f(x) using the graph of the function y=f(x) shown in the figure.

The graph of a function with a point removed

Solution.

When x goes to 1, the point (x,f(x)) on the graph goes to the empty dot whose y-coordinate is -1. So the limit limx1f(x)=1

Exercise 1.5 Use a table of functional values to evaluate limx3|x29|x3.

Solution.

To estimate the limit, we evaluate the function f(x)=|x29|x3 at some points that are close to 3, for example, x=30.01, 30.001, 30.0001, 3+0.0001, 3+0.001, 3+0.01. The function values can be seen in the following table.

x f(x) x f(x)
3-0.01 -5.99 3+0.01 6.01
3-0.001 -5.999 3+0.001 6.001
3-0.0001 -5.9999 3+0.0001 6.001

From the table we see that

6=limx3|x29|x3limx3+|x29|x3=6.

So the limit limx3|x29|x3 does not exist.

Exercise 1.6 Use a table of functional values to evaluate limx0cosx1x.

Solution.

From the following table,

x f(x) x f(x)
-0.01 0.00499996 0.01 -0.00499996
-0.001 0.0005 0.001 -0.0005
-0.0001 0.00005 0.0001 -0.00005

we estimate that limx0cosx1x=0.

Exercise 1.7 Consider the piece-wisely defined function f(x)={x1 if x<1x2+1 if x1.

  1. Evaluate the limits limx1f(x) and limx1+f(x).

  2. Determine if the limit limx1f(x) exists. Evaluate the limit if it exists. Otherwise, explain why it does not exist.

Solution.

Using a calculator or a software, we may graph the function. A part of the graph of the function is shown in the following figure (which is created using desmos).

The graph of a piecewise function

From the graph, we see that limx1f(x)=0 and limx1+f(x)=2.

So the limit limx1f(x) does not exist.

Exercise 1.8 Determine if the limit limx1|x+1|x+1 exists. Evaluate the limit if it exists. Otherwise, explain why it does not exist.

Solution.

From the graph of the function show below, we see that the left-limit and the right-limit do not agree. So the limit limx1|x+1|x+1 does not exist.

The graph of a function involve absolute value

Exercise 1.9 Determine the limit limx2x2+1x+2andlimx2+x2+1x+2.

Solution.

As x goes to 2, x2+1 goes to 5 and x+2 goes to 0. The limits go to either infinity or negative infinity.

When x<2, x+2<0 and x2+1x+2<0. So limx2x2+1x+2=.

When x>2, x+2>0 and x2+1x+2>0. So limx2x2+1x+2=.

This can also be seen by graphing the function.

The graph of a function with infinite limits at x=-2

Exercise 1.10 Determine the limit limx1+x2+xx2x2

Solution.

We note that the rational expression can be simplified and the limit equals limx1+x2+xx2x2=limx1+xx2.

Since xx2 goes to 112=13 as x goes to 1 from the right, the value of the limit is limx1+x2+xx2x2=13.

1.3 The Limit Laws

Theorem 1.1 (Basic Limit Results) For a constant number c, given a real number a, we have limxac=candlimxax=a.

Theorem 1.2 (The Limit Laws) Suppose that the limits limxaf(x)andlimxag(x) exists.
1. Sum: limxa(f(x)+g(x))=limxaf(x)+limxag(x). 2. Scalar Multiplication: For any constant c, limxa(cf(x))=climxaf(x). 3. Product: limxa(f(x)g(x))=limxaf(x)limxag(x). 4. Quotient: Suppose limxag(x)0. limxaf(x)g(x)=limxaf(x)limxag(x). 5. Power: For a positive integer n, limxa(f(x))n=(limxaf(x))n. 6. Radical: For a positive integer n, assume limxaf(x)0 if n is even, limxa(f(x)n=limxaf(x)n.

Theorem 1.3 (Substitution Theorem) Let P(x) and Q(x) be polynomials. Suppose that Q(a)0. Then limxaP(x)Q(x)=P(a)Q(a).

Theorem 1.4 (Squeeze Theorem) Suppose that the functions f and g are defined and f(x)h(x)g(x) holds on an interval (a,c)(c,b). If limxcf(x)=limxcg(x)=L, then limxch(x)=L. Note: The value c may be a finite number, , or . The quantity L may be a finite number, , or too.

Applying the squeeze theorem with |f(x)|f(x)|f(x)|, we get the following corollary.

Corollary 1.1 Suppose that limxa|f(x)|=0. Then limxaf(x)=0.

An important limit which can be obtained by using the squeeze theorem is the following one. limx0sinxx=1.

Exercise 1.11 Evaluate the limit limx22x21x3+2x2+1.

Solution.

Since (2)3+2(2)2+1=10, the limit can be evaluated by plugging in x=2. We get limx22x21x3+2x2+1=2(2)21(2)3+2(2)2+1=7.

Exercise 1.12 Evaluate the limit limx3(2x1)x+1.

Solution.

Since 3+1=4, the limit can be evaluated by plugging in x=3. We get limx3(2x1)x+1=2(3)1(3)+1=54.

Exercise 1.13 Evaluate the limit limt2(t22t33t+5)2

Solution.

By plugging in t=2, we get limt2t22t33t+5=2222332+5=27.

Applying the power rule, we get limt2(t22t33t+5)2=(27)=449.

Exercise 1.14 Evaluate the limit limh02x+h2xh.

Solution.

We cannot evaluate the limit directly by plugging in h=0. Because the denominator will be 0. However, you may notice that the plugging-in trick can be applied after rationalizing the numerator. Here is how to do that. limh02x+h2xh=limh0(2x+h2x)(2x+h+2x)h(2x+h+2x)=limh012x+h+2x=12x.

Exercise 1.15 Evaluate the limit limt0(1t1+t1t).

Solution.

Since directly plugging in t=0 does not work, we evaluate the limit by first do some algebraic operations (simplifying and rationalizing the numerator). limt0(1t1+t1t)=limt0(11+tt1+t)=limt0(1(1+t+1)1+t)=12

Exercise 1.16 Evaluate the limit limx2(3x2x21x2)

Solution.

We first simplify and then evaluate by plugging in. limx2(3x2x21x2)=limx2(3(x+1)x2x2)=limx2(2xx2x2)=limx2(1x+1)=13.

Exercise 1.17 Evaluate the limit limx0x2cos(1x).

Solution. Because x2ge0 and 1cos(1x)1. We see that x2x2cos(1x)x2. Note that limx0(x2)=0 and limx0x2=0. By the squeeze theorem, limx0x2cos(1x)=0.

Exercise 1.18 Evaluate the limit limx01sinx|x|.

Solution.

Note that 11sinx2 and |x|>0 for x0. Therefore 1|x|1sinx|x|2|x|.

Because limx01|x|=limx02|x|=. Then limx01sinxx=.

Exercise 1.19 Evaluate the limit limx0((x+1)2sinxx).

Solution. s

Since limx0sinxx=1 and limx0(x+1)2=(0+1)2=1. By the sum rule, limx0((x+1)2sinxx)=limx0(x+1)2limx0sinxx=11=0.

Exercise 1.20 Suppose limx0|f(x)|=0. Evaluate the limit limx0(f(x)1)2x2+2.

Solution.

Point-wisely, |f(x)|f(x)|f(x). Then by the squeeze theorem and the given assumption limx0|f(x)|=0, we can conclude that limx0f(x)=0.

By limit laws, we find that limx0(f(x)1)2x2+2=(limx0f(x)1)2limx0(x2+2)=(1)202+2=12.

1.4 Continuity

Intuitively, a function is continuous means that there is no holes or jumps when moving on the graph of the function.

Definition 1.1 A function f is continuous at x=a if limxaf(x)=f(a).

Note that the equality requires three true statement to hold.

  1. The function f is well-defined at a, that is a is in the domain of f.
  2. The limit limxaf(x) exists as a finite number.
  3. The limit equals the value f(a) of the function.

If f is defined near a but f is not continuous at a, we say that f is discontinuous at a.

The discontinuity is called a jumping discontinuity if the one-side limits exist but have different values.

A discontinuity is called a removable discontinuity if the limit exists.

A discontinuity is called an infinite discontinuity if a one-side limit is the infinity or the negative infinity.

Note that we may also define continuous from the left or the right using one-side limits.

A function f is continuous on an interval if it is continuous at every number in the interval, where at an endpoint of the interval, we understand continuity as from the left or the right.).

Derived from limit laws, we have the following theorem for continuity.

Theorem 1.5 Suppose that two function f and g are continuous at a. For a constant number c, the sum f+g, the scalar multiplication cf, the product fg, and the quotient fg (given that g(a)0) are all continuous at a.

As a corollary to the direct substitution theorem, we know that all polynomial and rational functions are continuous over their domain.

Indeed, it can be proved that root functions and trigonometric functions are also continuous at every number in their domains.

One important operation for producing new functions is the composition. For continuous functions, as you may expect, continuity works well with composition.

Theorem 1.6 Let f be a function continuous at c and g be a function continuous at f(c). Then fg is continuous at c.

A slightly more general result is the following.

Theorem 1.7 Let f(x) be a function continuous at L and g be a function such that limxcg(x)=L. Then limxcf(g(x))=f(L).

Continuous functions have many important (actually fundamental) properties. One of them, which has been used in the test-point method to solve inequalities in some algebra books, is call the intermediate value theorem.

Theorem 1.8 (Intermediate Value Theorem) Let f be a function continuous on the interval [a,b]. If f(a)f(b)<0, then there exists a number c(a,b) such that f(c)=0.

Exercise 1.21 Find the domain of the function f(x)=x1x2+1 and use definition of continuity and limit laws to determine whether the function is continuous over its domain.

Solution.

Since x2+1>0, the domain of the function is (,). By limit laws, for any number a, limxaf(x)=limxax1x2+1=a1a2+1=f(a). So the function is continuous over (,).

Exercise 1.22 Find the domain of the function f(x)={x22x3x3x34x=3. Use definition of continuity and limit laws to determine whether the function is continuous over its domain.

Solution.

The domain of the function is (,). When x3, the function is a rational function and hence continuous at any where except x=3. By limit laws, limx3f(x)=limx3x22x3x3=limx3(x+1)=4=f(3). So the function is continuous over (,).

Exercise 1.23 Find the domain of the function f(x)={x2+1x<01x=0sinxxx>0. Use definition of continuity and limit laws to determine whether the function is continuous over its domain.

Solution.

The domain of the function is (,). By properties of continuous functions, the function is continuous except possibly at x=0. When x=0, the left and right limits are limx0f(x)=limx0(x2+1)=1andlimx0+f(x)=limx0+sinxx=1. It follows that limx0f(x)=1=f(0). So the function is continuous over (,).

Exercise 1.24 Find the discontinuity and determine its type for the function f(x)=|x|x.

Solution.

Removing the absolute sign, the function can be written as f(x)={1x>01x<0 So the function has a discontinuity at x=0. It’s a jumping discontinuity because limx0f(x)=11=limx0+f(x).

Exercise 1.25 Find the discontinuity and determine its type for the function f(x)=2x2x3x+1.

Solution.

This function is a rational function which is undefined only at $x=-1#. So a possible discontinuity is at x=1. To determine the type, we find the limit first. limx1f(x)=limx12x2x3x+1=limx1(2x3)=1. So the function has a removable discontinuity at x=1.

Exercise 1.26 Evaluate the limit limx2(x1)213x2.

Solution.

Since products and compositions of continuous functions are continuous, using properties of continuous functions, we have limx2(x1)213x2=(21)21322=4.

Exercise 1.27 Evaluate the limit limxπ2sin(xcos(x)).

Solution.

Since products and compositions of continuous functions are continuous, using properties of continuous functions, we have limxπ2sin(xcos(x))=sin(π2cos(π2))=1.

Exercise 1.28 Find the value of the constant c such that the function f(x)={cx3+3 if x<1x2cx if x1 is continuous over (,).

Solution.

The function is only possibly discontinuous at x=1. To make the function continuous at x=1, we first need the existence of the limit. Since limx1f(x)=c+3andlimx1+f(x)=1c.
The limit of f exists at x=1 if and only if c+3=1c. Solve for c, we get c=1 and limx1f(x)=2. Hence f(1)=limx1f(x)=2.

So the function is continuous over (,) if c=1.

Exercise 1.29 Suppose f and g are continuous functions over (,) such that f(2)=1, g(2)=4 and f(4)=3. Find the limit limx2f(g(x))+g(x)2f(x)g(x)1

Solution.

Because f, g and the function y=x are continuous. The composite functions f(g(x)) and g(x) are continuous. By arithmetic rules and the definition of continuous functions, we have note that y=f(g(x))+g(x)2f(x)g(x)1 is a continuous function too. Therefore, limx2f(g(x))+g(x)2f(x)g(x)1=f(g(2))+g(2)2f(2)g(2)1=f(4)+42141=57.

Exercise 1.30 Use the Intermediate Value Theorem to show that the equation x3cos(x)1=0 has a root in the interval (0, 2).

Solution.

Let f(x)=x3cos(x)1. As a linear combination of continuous functions, f is also continuous. Note that f(0)f(2)=(0cos01)(23cos21)=2(7cos2)<0.

The IVT theorem tells us that there exists a number c(0,2) such that f(c)=0. So the equation has a solution x=c in (0,2).