Calculus I: Limits and Continuity, Exams of Calculus

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Calculus I

Rao W. O.

Course description

Limits and continuity of functions. Differentiation of functions of a single variable, para- metric and implicit differentiation. Antiderivatives and applications to areas and volumes of revolution. Integration by substitution. Applications of differentiation, mean value theorems of differential calculus.

References

1. Calculus: A complete course by Robert A. Adams and Christopher Essex.

iv

Part 1

The First Part

XEQ 104

x g pxq ˘ 1. 0 2 .0000 00000 ˘ 0. 1 2 .7048 13829 ˘ 0. 01 2 .7181 45927 ˘ 0. 001 2 .7182 80469 ˘ 0. 0001 2 .7182 81815 ˘ 0. 00001 2 .7182 81828

´ 1 ´ 0. 5 0 0. 5

1. 5

1

1. 5

2

2. 5

y

x

Figure 1.2. The graph of g pxq on the interval r´ 1 , 1 s

Note that g p 0 q is not defined, however we observe that

lim xÑ 0

g pxq “ lim xÑ 0

`

1 ` x^2

˘ 1 {x 2 “ e “ 2 .71828 1828 45 90 45....

Definition 1.1.1. If f pxq is defined for all x near a, except possibly at a itself, and if we can ensure that f pxq is as close as we want to L by taking x close enough to a, but not equal to a, we say that f approaches the limit L as x approaches a, and we write

lim xÑa

f pxq “ L.

Example 1.1.3. Find

(a) (^) xlimÑa c (where c is a constant)

(b) lim xÑa

x.

Solution. In part (a) we want to determine what c approaches as x approaches a. The answer is c approaches c; it cant be any closer to c than being at c i.e.

lim xÑa

c “ c.

In part (b) we want to determine what x approaches as x approaches a. The answer is a, i.e.

lim xÑa

x “ a.

When the denominator of a rational function is zero at x “ a, the limit as x Ñ a if it exists can be evaluated by first simplifying the expression to a form where the denominator is not zero at x “ a.

XEQ 104

Example 1.1.4. (a) lim xÑ´ 2

x^2 x ´ 2 x^2 5 x ` 6

(b) lim xÑa

x

a x ´ a (c) lim xÑ 4

x ´ 2 x^2 ´ 16 Solution. Note that in each case, the denominator yields zero at the indicated values. Thus we evaluate the limits by first simplifying the expression so that the denominator does not yield zero at the indicated value of x.

(a) (^) xlimÑ´ 2

x^2 x ´ 2 x^2 5 x ` 6

“ (^) xlimÑ´ 2

px 2 q px ´ 1 q px 2 qpx ` 3 q

“ (^) xlimÑ´ 2

x ´ 1 x ` 3

(b) (^) xlimÑa

x

a x ´ a

“ (^) xlimÑa

a ´ x ax x ´ a

“ (^) xlimÑa ´

ax

a^2

(c)

lim xÑ 4

x ´ 2 x^2 ´ 16

“ lim xÑ 4

p

x ´ 2 qp

x 2 q px ´ 4 qpx 4 qp

x ` 2 q

“ lim xÑ 4

x ´ 4 px ´ 4 qpx ` 4 qp

x ` 2 q

“ lim xÑ 4

px ` 4 qp

x ` 2 q

Definition 1.1.2. If f pxq is defined on some open interval pb, aq extending to the left of a, and if we can ensure that f pxq is as close as we want to L by taking x to the left of a and close enough to a, then we say that f has left limit L at x “ a, and we write

lim xÑa´^

f pxq “ L.

If f pxq is defined on some open interval pa, bq extending to the right of a, and if we can ensure that f pxq is as close as we want to L by taking x to the right of a and close enough to a, then we say that f has right limit L at x “ a, and we write

lim xÑa`^

f pxq “ L.

Example 1.1.5. Consider the signum function

sgnpxq “

x |x|

1 , x ą 0

´ 1 , x ă 0

x

y

1

´ 1

Figure 1.3. The graph of sgnpxq

XEQ 104

Theorem 1.1.4 (The Squeeze Theorem). Suppose f pxq ď gpxq ď hpxq holds for all x in some open interval containing a, except possibly at x “ a itself. Suppose also that

lim xÑa

f pxq “ lim xÑa

hpxq “ L.

Then lim xÑa

f pxq “ L also. Similar statements hold for left and right limits.

Example 1.1.8. Given that 3 ´ x^2 ď upxq ď 3 ` x^2 for all x ‰ 0 , find (^) xlimÑa upxq.

Solution. Since lim xÑ 0

p 3 ´ x^2 q “ 3 and lim xÑ 0

p 3 ` x^2 q “ 3 , the Squeeze Theorem implies

that lim xÑ 0

upxq “ 3.

Example 1.1.9. Show that if (^) xlimÑa |f pxq| “ 0 then (^) xlimÑa f pxq “ 0.

Solution. Since ´|f pxq| ď f pxq ď |f pxq|, and ´|f pxq| and |f pxq| both have limit 0 as x Ñ 0 , so does f pxq by the Squeeze Theorem.

Exercise 1.1.

1. Evaluate the limit or explain why it does not exist

(a) lim xÑ 4

`

x^2 ´ 4 x ` 1

(b) lim xÑ 2

3 p 1 ´ xq p 2 ´ xq

(c) lim xÑ 3

x 3 x 6

(d) lim tÑ 4

t^2 4 ´ t

(e) (^) tlimÑ 0

t ? 4 ` t ´

4 ´ t

(f) lim xÑ 1

x^2 ´ 1 ? x ` 3 ´ 2

(g) lim xÑ 3

x^2 ´ 6 x ` 9 x^2 ´ 9

(h) lim hÑ 0

4 ` h ´ 2 h

2. Find the indicated one-sided limit or explain why it does not exist.

(a) lim xÑ 2 ´

2 ´ x

(b) lim xÑ 2 `

2 ´ x

(c) lim xÑa´

|x ´ a| x^2 ´ a^2 (d) lim xÑa`

|x ´ a| x^2 ´ a^2

3. If lim xÑ 2

f pxq ´ 5 x ´ 2

“ 3 find lim xÑ 2

f pxq.

4. If lim xÑ 0

f pxq x^2

“ ´ 2 find lim xÑ 0

f pxq and lim xÑ 0

f pxq x

5. For the function g pxq given in the figure below, find each of the following limits or explain why it does not exist.

(a) lim xÑ 1

gpxq (b) lim xÑ 2

gpxq (c) lim xÑ 3

gpxq

XEQ 104

1 2 3

y “ g pxq y

x

1.2. Limits at Infinity and Infinite Limits

1.2.1. Limits at Infinity. Consider the function f pxq “

x ? x^2 ` 1

whose graph is

x

y

1

´ 1

Figure 1.4. The graph of f pxq “? x x^2 ` 1

It can be verified by direct computation that lim xÑ8 f pxq “ 1 and (^) xÑ´8lim f pxq “ ´1.

Definition 1.2.1. If f pxq is defined on an interval pa, 8q and if we can ensure that f pxq is as close as we want to the number L by taking x large enough, then we say that f pxq approaches the limit L as x approaches infinity, and we write

lim xÑ

f pxq “ L.

If f pxq is defined on an interval p´8, bq and if we can ensure that f pxq is as close as we want to the number M by taking x negative and large enough in absolute value, then we say that f pxq approaches the limit M as x approaches negative infinity, and we write

xÑ´8^ lim f^ pxq “^ M.

Example 1.2.1.

Consider the function f pxq “

x

whose graph is given below

XEQ 104

In general, let Pmpxq “ amxm^ ¨ ¨ ¨ a 0 and Qnpxq “ bnxn^ ¨ ¨ ¨ b 0 be polynomilas of degrees m and n respectively, so that am ‰ 0 and bn ‰ 0. Then

xÑ˘8lim

Pmpxq Qnpxq

0 , if m ă n am bn

, if m “ n

does not exist, if m ą n.

1.2.2. Infinite Limits. Here we investigate the behaviour of functions as image points move infinitely far away from the origin.

Example 1.2.5.

Describe the behaviour of f pxq “

x^2

near x “ 0.

Solution. Note that f p 0 q is not defined. However lim xÑ 0

f pxq “ lim xÑ 0

x^2

x

y

y “ (^) x^12

Figure 1.6. The graph of f pxq “ 1 x^2

Example 1.2.6.

Describe the behaviour of f pxq “

x

near x “ 0.

Solution. lim xÑ 0 ´^

f pxq “ lim xÑ 0 ´

x

“ ´8 and lim xÑ 0 `^

f pxq “ lim xÑ 0 `

x

“ 8. The figure

illustrates these results.

x

y

p 1 , 1 q

p´ 1 , ´ 1 q

Figure 1.7. The graph of f pxq “

1 x

XEQ 104

Example 1.2.7.

(a) (^) xlimÑ8p 3 x^3 ´ x^2 ` 2 q “ 8 (b) lim xÑ

px^4 ´ 5 x^3 ´ xq “ 8

(c) (^) xÑ´8lim p 3 x^3 ´ x^2 ` 2 q “ ´ (d) (^) xlimÑ8px^4 ´ 5 x^3 ´ xq “ 8

The highest-degree term of a polynomial dominates the other terms as |x| grows larger, so that the limit of this term at ˘8 determine the limit of the whole polynomial. For example observe that

3 x^3 ´ x^2 ` 2 “ 3 x^2

3 x

`

3 x^3

The factor in the parentheses approaches 1 as x approaches ˘8, so the behaviour of the polynomial is just that of its highest-degree term 3x^3.

Example 1.2.8.

Evaluate (^) xlimÑ

x^3 1 x^2 1

Solution.

x^ limÑ

x^3 1 x^2 1

“ (^) xlimÑ

x `

x^2 1 `

x^2

Exercise 1.2.

1. Evaluate the limits in

(a) (^) xlimÑ

x 2 x ´ 3

(b) lim xÑ

x x^2 ´ 4

(c) lim xÑ

3 x^3 ´ 5 x^2 7 8 2 x ´ 5 x^3

(d) lim xÑ´

x^2 ´ 2 x ´ x^2 (e) lim xÑ´

x^2 3 x^3 2 (f) (^) xlimÑ

x^2 sin x x^2 cos x (g) lim xÑ

3 x ` 2

x 1 ´ x

2. Evaluate the indicated the limit. If it does not exist, is the limit 8 , ´8, or neither?.

(a) lim xÑ 3

3 ´ x

(b) lim xÑ 3

p 3 ´ xq^2

(c) lim xÑ´ 5 { 2

2 x 5 5 x 2 (d) lim xÑ´ 2 { 5

2 x 5 5 x 2

1.3. Continuity Definition 1.3.1. A function f is continuous at an interior point c of its domain if

x^ limÑc f^ pxq “^ f^ pcq. If either lim xÑc f pxq fails to exists or it exists but is not equal to f pcq, then we say that

f is discontinuous at c. The following graphs illustrate what is described in the definition above.

XEQ 104

(d) All rational functions are continuous wherever they are defined. (e) All absolute value functions are continuous wherever they are defined. (f) The sine, cosine, tangent, secant, cosecant and cotangent functions are continuous wherever they are defined.

Theorem 1.3.2. If the functions f and g are both defined on an interval containing c and both are contin- uous at c, then the following functions are also continuous at c.

1. f ˘ g
2. f g
3. kf , k is any number.
   4. f {g provided gpcq ‰ 0.
   5. pf pxqq^1 {n, provided f pxq ą 0 if n is even.

Theorem 1.3.3. IIf f pgpxqq is defined on an interval containing c and f is continuous at L and lim xÑc

gpxq “ L,

then

lim xÑc

f pgpxqq “ f pLq “ f

lim xÑc

gpxq

In particular, if g is continuous at c (so that L “ gpcq), then the composition f ˝ g is continuous at c and

lim xÑc

f pgpxqq “ f pgpcqq.

Continuous Extensions and Removable Discontinuities If f pcq is not defined, but lim xÑc f pxq “ L exists, we can define a new function F pxq by

F pxq “

f pxq, if x is in the domain of f L, ifx “ c.

F pxq is called a continuous extension of f pxq to c.

Example 1.3.3.

Show that f pxq “

x^2 ´ x x^2 ´ 1

has a continuous extension at x “ 1 and find that extension.

Solution. Observe that f pxq is not defined. However

f pxq “

x^2 ´ x x^2 ´ 1

xpx ´ 1 q px ´ 1 qpx ` 1 q

x x ` 1

. 6 lim xÑ 1

f pxq “ lim xÑ 1

x x ` 1

The function F pxq “

x x ` 1

is equal to f pxq for x ‰ 1 and is also continuous at x “ 1 ,

where F p 1 q “

. Thus the continuous extension of f pxq to x “ 1 is F pxq. Below are the

curves of f pxq and F pxq.

XEQ 104

x

y y “ x

(^2) ´x x^2 ´ 1

(a)

x

y

y “ (^) xx` 1

(b)

If a function is undefined o discontinuous at a point c but can be redefined at that single point so that it becomes continuous there, then we say that f has a removable discontinuity at at x “ c.

Example 1.3.4.

The function gpxq “

x, x ` 2

1 , x “ 2

has a removable discontinuity at x “ 2. To remove it

redefine gp 2 q “ 2. The corresponding graph is given below

x

y

y “ gpxq

p 2 , 1 q

Figure 1.

Exercise 1.3.

1. State where in its domain the given function is continuous, where it is left or right continuous, and where it is just discontinuous.

(a) f pxq “

x if x ă 0

x^2 if x ě 0.

(b) f pxq “

x if x ă 1

x^2 if x ě ´ 1.

(c) f pxq “

1 {x^2 if x ‰ 0

0 if x “ 0.

(d) f pxq “

x^2 if x ď 1

1. 987 if x ą 1.

CHAPTER 2

Derivatives

2.1. Tangent Lines and Their Slopes

Consider a graph C of a function which is continuous on an interval containing a point x 0. Let P px 0 , f px 0 qq and Qpx 0 h, f px 0 hqq where h ą 0 be points on the curve.

x

y

x (^0) x 0 ` h

L

y “ f pxq

P

Q

Figure 2.

The line passing through P and Q is called a secant line to the curve. This line rotates about P as Q moves along the curve. If L is the line whose slope is the limit of the slopes of these secant lines P Q as Q approaches P along C, then L is the tangent to C at P. The slope of P Q is

f px 0 ` hq ´ f px 0 q h

This expression is called the Newton quotient or difference quotient for f at x 0.

Definition 2.1.1 (Non vertical tangent lines). Suppose that the function f is contin- uous at x 0 and that

lim hÑ 0

f px 0 ` hq ´ f px 0 q h

“ m

exists. Then the straight line having slope m and passing through the point P px 0 , f px 0 qq is called the tangent line (or simply the tangent) to the graph of y “ f pxq at P. The equation of the tangent line is

y “ mpx ´ x 0 q ` yq.

Example 2.1.1. Find an equation of the tangent line to the curve y “ x^2 at the point p 1 , 1 q.

XEQ 104

Solution. Here f pxq “ x^2 , x 0 “ 1 , y 0 “ 1. Slope of the tangent line is

m “ lim hÑ 0

f p 1 ` hq ´ f p 1 q h

“ lim hÑ 0

p 1 ` hq^2 ´ 1 h

“ lim hÑ 0

h^2 ` 2 h h

“ lim hÑ 0

p 2 ` hq “ 2.

Accordingly the equation of the tangent line is y “ 2 px ´ 1 q ` 1 , that is, y “ 2 x ´ 1.

x

y y “ x^2

y “ 2 x ´ 1

p 1 , 1 q

Figure 2.

Definition 2.1.2 (Vertical tangent lines). If f is continuous at P px 0 , f px 0 qq and if either

lim hÑ 0

f px 0 ` hq ´ f px 0 q h

then the vertical line x “ x 0 is tangent to the graph y “ f pxq at P. If the limit of the Newton quotient fails to exist in any other way other than being ˘8, then the graph y “ f pxq has no tangent line at P.

Example 2.1.2. Determine the tangent lines of the following curves at x “ 0.

(a) f pxq “ 3

x (b) f pxq “ x^2 {^3 (c) f pxq “ |x|.

Solution.

(a)

f p 0 ` hq ´ f p 0 q h

h^1 {^3 ´ 0 h

h^2 {^3

ñ lim hÑ 0

h^2 {^3

It follows that the tangent line to y “ x^1 {^3 at x “ 0 is the line x “ 0.

(b)

f p 0 ` hq ´ f p 0 q h

h^2 {^3 ´ 0 h

h^1 {^3

. Note that

lim hÑ 0 `

h^1 {^3

“ 8 while lim hÑ 0 ´

h^1 {^3

and so the limit does not exist. It follows that the graph pf y “ x^2 {^3 does not have a tangent line at x “ 0.

x

y y “ x^2 {^3

Figure 2.