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COURSE NOTES MATH 100: PROF. DRAGOS GHIOCA

http://www.math.ubc.ca/˜dghioca/courses/100.html

1. September 8 1.1. Some of the things you need to know from high school. • basic trigonometric identities and formulas such as: π  π   π  √3  π 1 cos(0) = 1, sin = √ , ······ , cos = 0, sin = 1, cos − = 2 4 2 2 3 2 sin2 (x) + cos2 (x) = 1

and thus, −1 ≤ sin(x), cos(x) ≤ 1. • graphs for basic functions such as: x2 , sin(x), ln(x), ex , but also for piecewise defined functions such as  7 − x if x < 2 f (x) = 2x + 3 if x ≥ 2 • the slope of the line passing through two points (x1 , y1 ) and (x2 , y2 ) is m :=

y2 − y1 . x2 − x1

• equation of the line passing through a point (x0 , y0 ) and having slope m is y − y0 = m(x − x0 ).

• basic algebra such as a2 − b2 = (a − b)(a + b). You can read more about all these topics, using the CLP textbook – see Chapter 0 The basics and Appendix A High school material – both chapters contain even more information than what you need to know in order to take this course. 1.2. Tangent lines seen as a limit case for the secant lines. Given the function f (x) = x2 , the secant line corresponding to the two points (3, 9) and (5, 25) on this parabola has slope 25 − 9 16 m= = =8 5−3 2 and so, the secant line has the equation y − 9 = 8 · (x − 3), which can be simplified to y = 8x − 15. If we fix the point (3, 9) and only vary the second point on the parabola, each time we get a different secant line. For example, let’s consider the case when the 1

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MATH 100: PROF. DRAGOS GHIOCA

second point for the secant line is an arbitrary point (x, x2 ) on the given parabola. Then the secant line determined by the points (3, 9) and (x, x2 ) is x2 − 9 (x − 3)(x + 3) = x + 3. = x−3 x−3 Note that the slope of this line depends on x (hence also its notation mx ) which makes sense since the slope of the secant line changes with the choice of the second point (x, x2 ). Now, if we want to find the slope of the tangent line at the point (3, 9) on the graph of the parabola y = x2 , then we can view this slope of the tangent line as the slope of a secant line between the points (3, 9) and (x, x2 ) where the second point approaches the first point, i.e., x approaches 3. With this understanding, we see then that the tangent line at the point (3, 9) on the given parabola has slope mtg which is obtained from the slope of the secant line mx :=

mx = x + 3, as we let x approaches 3, which means that the tangent line has slope mtg = 3 + 3 = 6. In conclusion, the equation of the tangent line at the point (3, 9) on the parabola y = x2 is y − 9 = 6(x − 3), i.e., y = 6x − 9. For more information regarding tangents, see Section 1.1 in the CLP textbook Drawing tangents and a first limit. You should also read Section 1.2 Another limit and computing velocity; important concepts to know later are the ones for average velocity and also for instantaneous velocity. 2. September 10 2.1. Definition of the limit. Definition 2.1. For a function f (x) and a real number a, we say that f (x) has limit L (which is also a real number) as x approaches a and we denote this by lim f (x) = L

x→a

if the values of f (x) get closer and closer to L as x gets closer and closer to a. 2.2. Examples. • limx→2 x2 = 4 • limx→ π2 sin(x) = 1 √ • limx→9 x = 3 1 • limx→−8 x 3 = −2 • limx→1 ln(x) = 0. Also, if f (x) = then



cos(3x) if x 6= π6 5 if x = π6

limπ f (x) = limπ cos(3x) = cos

x→ 6

x→

6

π  2

= 0;

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3

in other words, it doesn’t matter what is the actual value of f (x) at x = π6 ; instead, all it matters is the tendency of the function f (x) as x approaches 6π and  clearly  π since cos(3x) approaches cos 3 · π the function tends to 0 as x approaches 6 6 = π  cos 2 = 0. 2.3. Lateral limits. For the function  7−x f (x) = 2x + 3

we have that

if x < 2 if x ≥ 2

lim f (x) does not exist

x→2

since f (x) has different tendencies as you approach x = 2. From the left, the tendency for f (x) is to approach 7 − 2 = 5 since the definition of the function for x < 2 is f (x) = 7 − x. On the other hand, from the right (as you approach x = 2), the tendency of the function is to approach 2 · 2 + 3 = 7 since the definition of the function for x > 2 is f (x) = 2x + 3. Since 5 6= 7, we do not have the same tendency for f (x) as you approach x = 2 from both left and right. Once again, as in the previous example, the actual value of f (x) at x = 2 (which is f (2) = 7 according to the definition of f (x)) is completely irrelevant for the computation of the limit. The above example motivates the definition of the left and right limits (as x approaches 2) since the tendencies of f (x) both from the left and from the right are well-defined, i.e., lim− f (x) = lim− 7 − x = 5 x→2

x→2

and lim f (x) = lim+ 2x + 3 = 7,

x→2+

x→2

where we used the following definition for lateral limits: Definition 2.2. For a function f (x) and a real number a, we write lim f (x) = L

x→a−

if the values of f (x) get closer and closer to the real number L as x gets closer and closer to a from the left, i.e., when x → a but x < a. For a function f (x) and a real number a, we write lim f (x) = L

x→a+

if the values of f (x) get closer and closer to the real number L as x gets closer and closer to a from the right, i.e., when x → a but x > a. 2.4. Infinite limits.

1 does not exist x since the values of x1 become larger and larger (and positive) as x becomes smaller and smaller (and positive). However, even if the limit doesn’t exist in this case but since there is a clear tendency for x1 as x → 0+ , we write that lim

x→0+

lim

1

x→0+ x

diverges to + ∞.

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MATH 100: PROF. DRAGOS GHIOCA

Similarly, we write that 1 diverges to − ∞ x because the values of x1 become larger and larger (but negative), as x becomes smaller and smaller (but negative). More generally, we have the definitions: lim

x→0−

Definition 2.3. We write lim f (x) diverges to + ∞

x→a

if the values of f (x) become larger and larger and also positive, as x approaches a. Similarly, we write lim f (x) diverges to − ∞ x→a

if the values of f (x) become larger and larger and also negative, as x approaches a. Finally, we also use lim± f (x) diverges to ± ∞ x→a

if the values of f (x) become larger and larger and they are all positive (or all negative), as x approaches a (either from the right, or from the left). If limx→a+ f (x) diverges to +∞ or it diverges to −∞, or if limx→a− f (x) diverges to +∞ or it diverges to −∞, then we say that x = a is a vertical asymptote (V.A.) for f (x). 1 The classical example of a vertical asymptote would be x−2 which has a vertical asymptote at x = 2; we check this by verifying one of the two lateral limits, say: 1 diverges to + ∞. x−2 So, x = 2 is a V.A. for f (x). We don’t need to check also the left limit at x = 2; one lateral limit divergent to +∞ or −∞ suffices. Also, very important to have in mind the following example of a vertical asymptote that even touches the graph of the function.  1 if x < 0 x f (x) = 5 if x ≥ 0 lim

x→2+

In this case, we see that x = 0 is a V.A. for f (x) simply because 1 diverges to − ∞. x It doesn’t mater that the limit from the right at x = 0 is convergent: lim f (x) = lim

x→0−

x→0−

lim f (x) = lim 5 = 5; +

x→0+

x→0

we still have a V.A. at x = 0. Additional examples: 5 5 • limx→1− x−1 diverges to −∞. So, the function x−1 has a vertical asymptote x = 1. −3 −3 • limx→2+ 4−x 2 diverges to +∞. So, the function 4−x2 has a vertical asymptote at x = 2. This is not the only vertical asymptote for the function as −3 lim − diverges to + ∞ x→−2 4 − x2

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5

since for x approaching −2 from the left, the denominator of the above fraction becomes arbitrarily small, but negative. This shows that also x = −3 −2 is a vertical asymptote for 4−x 2. 1 • limx→0 x2 diverges to +∞. • limx→ π2 tan(x) does not exist, but we cannot say where exactly it diverges since limπ tan(x) diverges to + ∞ x→ 2 −

while lim tan(x) diverges to − ∞.

x→ 2π +

You can read more about limits in Section 1.3 The limit of a function from the CLP textbook. 3. September 13 3.1. Limit laws. We know that limx→a f (x) = L1 and limx→a g(x) = L2 . Then • • • •

limx→a c · f (x) = c · L1 for any real number c. limx→a f (x) ± g(x) = L1 ± L2 . limx→a f (x) · g(x) = L1 · L2 . f (x) limx→a g(x) = LL12 if L2 6= 0. On the other hand, if L2 = 0 but L1 6= 0, then lim

x→a

L1 f (x) = doesn’t exist L2 g(x)

and if we have more information about g(x) (such as what is the sign of g(x) as x approaches a), then we could even determine whether the above f (x) limit of g(x) actually diverges to either +∞ or −∞, or perhaps it simply

f (x) doesn’t exist (in case g(x) takes values arbitrarily large and positive but also arbitrarily large and negative). • limx→a fp (x)n = L1n for any positive integer n. √ n f (x) = n L1 as long as L1 ≥ 0 when n is even (and also assuming • lim x→a p n f (x) makes sense as a function as x approaches a). If n is odd, there is no restriction on f (x) (nor on L1 ).

Examples: • • • •

√ √ limx→5 x + 4 = 5 + 4 = 3 2 2 +1 = 2 limx→3 xx3+1 = 333 −2 5 −2 x 0 limx→0 e = e = 1 limx→1 ln(x) = ln(1) = 0

3.2. Additional methods for computing limits. 2

−2x • limx→2 x4−x 2 In order to compute this limit, which is of the form “ 00 ”, we need to factor out (both numerator and denominator) and then cancel out the comon factor from the denominator and the numerator. So,

x2 − 2x 1 2 x x(x − 2) =− . = = lim = lim x→2 −(2 + x) x→2 (2 − x)(2 + x) x→2 4 − x2 −4 2 lim

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MATH 100: PROF. DRAGOS GHIOCA √

−2 • limt→3 t+1 t−3 This time we need to multiply both the numerator and the denominator √ by the conjugate of the numerator, which is t + 1 + 2 and then factor out the common factor (t − 3) which will also appear in the numerator. So, we get √   √ √ t+1−2 · t+1+2 t+1−2 lim = lim √  t→3 t→3 t−3 (t − 3) t + 1 + 2 (t + 1) − 4 √  = lim t→3 (t − 3) t+1+2 t−3 √  = lim t→3 (t − 3) t+1+2 1 = lim √ t→3 t+1+2 1 = . 4 1

• limx→0− 2 x In this case, we analyze from the inside the given function and so, we see that x1 goes to (or moreover diverges to) −∞ as x approaches 0 from the left. So, in the above limit, as x approaches 0 from the left, we compute something of the form “2large and negative number ” which means that  the above limit is equal to 0. • limx→0+ cos 1x In this case, as x approaches 0 (from the right), we have that 1x diverges 1  (to +∞), which means that cos x cannot approach a given real number (instead, it simply oscillates between −1 and 1); thus   1 lim+ cos does not exist. x→0 x   • limx→0 x2 sin 1x   As in the previous case, we have that sin 1x does not approach a given real number as x approaches 0; instead, it simply oscillates between −1 and 1. However, as opposed to the previous example, we multiply everything by x2 which converges to 0 as x approaches 0. So, we multiply a number very close to 0 by a number which is always in between −1 and 1; so, we expect the limit to be equal to 0. We prove formally our expectation as follows:   1 −1 ≤ sin ≤1 x for x approaching 0 (actually, in this case, for all real numbers x 6= 0) and so, multiplying the above inequalities by x2 (which is positive), we get   1 ≤ x2 . −x2 ≤ x2 sin x

COURSE NOTES

7

2 2 However, lim  1x→0  −x = 0 = limx→0 x , which means that the middle func2 tion x sin x is squeezed to 0 as x approaches 0; hence   1 2 = 0. lim x sin x→0 x

This is justified through the use of the so-called Squeeze Theorem. Theorem 3.1. (Squeeze Theorem) If g(x) ≤ f (x) ≤ h(x)

for all x in a neighborhood of the real number a, and if lim g(x) = L = lim h(x), x→a

x→a

then also limx→a f (x) = L. You can read more about limits in Section 1.4 Calculating limits with limit laws from the CLP textbook. 4. September 15 Definition 4.1. If the values taken by the function f (x) get closer and closer to the real number L as x becomes larger and larger and also positive, then we say that the limit of f (x) equals L as x approaches +∞, and we write lim f (x) = L.

x→+∞

Similarly, if the values taken by the function f (x) get closer and closer to the real number L as x becomes larger and larger and also negative, then we say that the limit of f (x) equals L as x approaches −∞, and we write lim f (x) = L.

x→−∞

When limx→+∞ f (x) = L or limx→−∞ f (x) = L, then y = L is a horizontal asymptote (H.A.) for f (x). 4.1. Examples. • limx→+∞ x4 = 0 because 4x becomes arbitrarily small as x becomes large. So, the function 4x has the horizontal asymptote y = 0.  1 approaches 0 as x gets large and cos(0) = 1. • limx→+∞ cos 1x = 1 because x 1  So, the function cos x has the horizontal asymptote y = 1. 2 +2x+6 • limx→+∞ 5x = 35 because after dividing both numerator and denom3x2 −x+7 inator by the highest power of x appearing in the denominator, we get 5x2 +2x+6 x2 3x2 −x+7 x2 5 + x2 + x62 lim x→+∞ 3 − 1 + 72 x x

5x2 + 2x + 6 = lim x→+∞ 3x2 − x + 7 x→+∞ lim

= = because

2 , 6 , 1 x x2 x

and

7 x2

5 , 3

all converge to 0 as x approaches +∞.

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MATH 100: PROF. DRAGOS GHIOCA √

2

• limx→+∞ 2x x+5+3 We divide again both denominator and numerator by the highest power of the denominator and so, r r √ √ 2x2 + 5 3 5 3 2x2 + 5 + 3 lim 2 + 2 + = 2. + = lim = lim 2 x→+∞ x→+∞ x→+∞ x x x x x √ 2 • limx→−∞ x + x + x The above limit is of the form “∞ − ∞” as x approaches −∞ and this is an√indeterminate form. We first need to multiply by the conjugate, which is x2 + x − x and therefore, we get √ √ p ( x2 + x + x)( x2 + x − x) x2 + x + x = lim lim √ x→−∞ x→−∞ x2 + x − x 2 x + x − x2 = lim √ x→−∞ x2 + x − x x . = lim √ 2 x→−∞ x +x−x In the last limit, the denominator is of the form “∞ + ∞” as x approaches −∞. So, we divide both numerator and the denominator of the above fraction by the highest power of x appearing in the denominator, which is x. So, 1 x lim √ = lim √x2 +x 2 x→−∞ x→−∞ x +x−x −1 x and then we note that r r √ x2 + x 1 x2 + x =− 1+ =− x x2 x

because x < 0 as x approaches −∞ (while a square-root is always positive). So, we conclude that p x x2 + x + x = lim √ 2 lim x→−∞ x→−∞ x +x−x 1 = lim q x→−∞ − 1 + x1 − 1 =

1 1 =− . −1 − 1 2

• limx→+∞ cos(x) x As x approaches ∞, cos(x) oscillates between −1 and 1 and therefore, the numerator by itself has no limit; however, since we divide each time a number between −1 and 1 by a number x approaching ∞, we expect the above limit equals 0. Indeed, this is formalized as follows: −1 ≤ cos(x) ≤ 1 and so, after dividing by x, we have −

1 1 cos(x) ≤ ≤ x x x

COURSE NOTES

and therefore, because limx→+∞ − x1 = 0 = limx→+∞ squeezed to 0, i.e., cos(x) = 0, lim x→+∞ x by Squeeze Theorem.

9 1 , x

then our limit is

Theorem 4.2. (Squeeze Theorem) If g(x) ≤ f (x) ≤ h(x) for all x > M (where M is some given real number, possibly very large and positive), and if lim

x→+∞

g(x) = L = lim h(x), x→+∞

then also limx→+∞ f (x) = L. Similarly, if g(x) ≤ f (x) ≤ h(x) for all x < M (where M is some given real number, possibly very large and negative), and if lim

x→−∞

g(x) = L = lim h(x), x→−∞

then also limx→−∞ f (x) = L. You can read more about limits at infinity in Section 1.5 Limits at infinity from the CLP textbook. 5. September 17 Definition 5.1. We say that f (x) is continuous at x = a if limx→a f (x) = f (a). So, in particular, for f (x) to be continuous at x = a, we need implicitly 3 things: • limx→a f (x) exists and it equals some real number L; • f (x) is defined at x = a; and • L = f (a). Polynomials, trigonometric, exponentials, logarithmic functions, along with sums, products, quotients and compositions of such functions are continuous on their domain of definition. 2 +2 For example, f (x) = xx−2 is continuous for all real numbers x except for x = 2 since at x = 2 the function is not even defined; but everywhere else, the function is a quotient of two polynomials and so, it’s continuous on R \ {2} = (−∞, 2)∪ (2, +∞). Another example is g(x) = tan(x) which is continuous whenever the denominator sin(x) of cos(x) is not equal to 0 and so, tan(x) is continous for all x which is not of the form 2π + kπ where k is an integer, i.e., g(x) is continuous on o nπ + kπ : k ∈ Z . R\ 2 √ The function h(x) = 5 − x is continuous whenever is well-defined, which is equivalent with 5 − x ≥ 0, i.e., h(x) is continuous on (−∞, 5]. Now, at x = 5, we note that the limit limx→5 h(x) makes sense only from the left since the domain of h(x) is all real numbers x ≤ 5. This motivates also the following definitions. Definition 5.2. We say that f (x) is continous from the left at x = a (or leftcontinuous at x = a) if limx→a− f (x) = f (a). Similarly, we say that f (x) is continous from the right at x = a (or rightcontinuous at x = a) if limx→a+ f (x) = f (a).

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MATH 100: PROF. DRAGOS GHIOCA

√ So, for the function 5 − x, technically speaking the function is only continuous from the left at x = 5 but since in this case, the function is not defined to the right of x = 5, then – for simplicity – we simply say that the function is continuous at x = 5. However, when the function f (x) is defined from both sides of some number x = a and the function f (x) is given by a different formula on the left, respectively on the right of x = a, then it is mandatory that we check both left and right limits of f (x) at x = a and we note that in this case f (x) is continuous at x = a if and only if lim− f (x) = f (a) = lim+ f (x). x→a

x→a

In other words, if just one of the two lateral limits above do not exist (or are not equal to f (a)), then f (x) would fail to be continuous at x = a. 5.1. Examples. • The function f (x) =



3 − x2 if x ≤ 2 x + 1 if x > 2

is not continuous at x = 2; we have a jump discontinuity at x = 2. Indeed, we compute lim f (x) = lim− 3 − x2 = 3 − 22 = −1

x→2−

x→2

which is not equal to lim f (x) = lim+ x + 1 = 2 + 1 = 3;

x→2+

x→2

so, limx→2 f (x) doesn’t exist. Therefore, f (x) is not continuous at x = 2. On the other hand, since f (2) = −1, we see that f (x) is continuous from the left at x = 2 (but it’s not continuous from the right at x = 2). Also, everywhere else away from the branch point x = 2, f (x) is continuous since for x < 2, f (x) is simply a quadratic polynomial (so, it’s continuous), while for x > 2, f (x) is a linear polyomial (again continuous). So, f (x) is continuous on R \ {2} = (−∞, 2) ∪ (2, +∞). • The function  sin(x) if x 6= 6π f (x) = 3 if x = 6π

is not continuous at x = π6 ; we have a removable discontinuity at x = π6 . Indeed, this time, we have that limx→ π6 f (x) exists and it is equal to π 1 limπ f (x) = limπ sin(x) = sin = ; x→ 6 x→ 6 6 2 π  however, this value does not match f 6 = 3 – this is why f (x) is not continuous at x = π6 (in this case, f (x) is neither continuous from the left nor from the right at x = π6 ). On the other hand, everywhere else, away from x = π6 , f (x) is continuous since it’s which is continuous; so, f (x) is continuous   ...