This Week Sections 2.1-2.3,2.5,2.6 First homework due Tuesday night at 11:30 p.m. Average and instantaneous velocity worksheet Tuesday available at http://www.math.washington.edu/ m124/ (under week 2) print it out before coming to class
Velocity If an arrow is shot upward on the moon with a velocity of 58 meters per second. It height in meters is given by p(t) = 58t 0.83t 2. 1 Find the average velocity over the time intervals [1, 2], [1, 1.5], [1, 1.1], [1, 1.01], and [1, 1.001]. 2 Find the instantaneous velocity at t = 1.
Average Velocity between 1 and t Average velocity is change in position divided by change in time. Average velocity between time 1 and t is p(t) p(1). t 1 p(t) h(1) t 1 t 2 55.51 1.5 55.925 1.1 56.257 1.01 56.3317 1.001 56.339
Instantaneous Velocity To see what happens as h approaches 0 we do a little algebra. p(1 + h) p(1) (1 + h) 1 = 58(1 + h) 0.83(1 + h)2 (58 0.83(1) 2 ) h = 58 + 58h 0.83(1 + 2h + h2 ) (58 0.83) h = 58h 1.66h 0.83h2 ) h = 56.34 0.83h As h approaches 0 the average velocity approaches 56.34. We say the instantaneous velocity at t = 1 is 56.34 and write p(1 + h) p(1) Instantaneous velocity = lim = 56.34. h 0 h
Geometric Interpretation p(1+h) p(1) h is the slope of the secant line between (1, p(1)) and (1 + h, p(1 + h)). As h approaches 0 the secant lines approach the tangent line at (1, p(1)). The instantaneous velocity of 56.34 is the slope of the tangent line at (1, p(1)).
Two Interpretations All of our work over the next few weeks can be interpreted as finding methods to determine the instantaneous velocity for an arbitrary position function. Geometrically, our work can be interpreted as determining the slope of a tangent line for an arbitrary point on the graph of an arbitrary function. In order to do this we first must discuss what we mean by the limit of a function as we approach a point.
Limits lim x a f (x) = L means that we can make the value of f (x) arbitrarily close to L by taking x sufficiently close to a but not equal to a. In this picture lim x 2 f (x) = 4.
lim x a f (x) does not depend on the value of f (a). In all of these pictures lim x 0 f (x) = 1. For most of the limits lim x a f (x) = L that we take in this course f (a) will not be defined.
Naive Idea We will plug in small values of x to try to find lim x 0 sin(x). x sin(x) x x ±1.84147 ±1/2.95885 ±.4.973545 ±.3.985067 ±.001.99999 Based on this we might be tempted to say lim x 0 sin(x)/x = 1.
Graphical Interpretation Looking at the graph of y = sin x/x we can see that lim sin(x)/x = 1. x 0
Precise Definition of a Limit The precise definition of a limit is contained in Section 2.4. We won t cover this section in class. We will be content to know limits when we see them.
Problematic Example Find lim sin(π/x). x 0 sin(π) x x ±1 0 ±1/2 0 ±1/3 0 ±1/10 0 ±1/100 0 ±1/n 0 Based on this we might be tempted to say lim sin(π/x) = 0. x 0
But if we would have chosen x = 2 1 + 4n then we would have seen a different behavior as the points approached 0. For these values ( π ) sin(π/x) = sin 2 + 2nπ = 1. Also if we would have chosen x = 2 3 + 4n then we would have sin(π/x) = sin ( 3π 2 + 2nπ) = 1.
sin(π/x) oscillates between 1 and 1 infinitely often as x 0. Thus lim x 0 sin(π/x) does not exist.
Check your work On every limit calculation where you want to find lim f (x) x a you should plug in values of x near a. If the numbers you get are not close to your answer then you did something wrong.
Calculating limits graphically
Calculating limits graphically For the function h whose graph is given, state the value of each quantity if it exists, If it does not exist, explain why 1 lim x 3 h(x) 2 lim x 3 + h(x) 3 lim x 3 h(x) 4 h( 3) 5 lim x 0 h(x) 6 h(0) 7 lim x 2 h(x) 8 h(2) 9 lim x 5 + h(x) 10 lim x 5 h(x)
Calculating limits graphically For the function h whose graph is given, state the value of each quantity if it exists, If it does not exist, explain why 1 lim x 3 h(x) = 4 2 lim x 3 + h(x) = 4 3 lim x 3 h(x) = 4 4 h( 3) does not exist 5 lim x 0 h(x) does not exist 6 h(0) = 1 7 lim x 2 h(x) = 2 8 h(2) does not exist 9 lim x 5 + h(x) = 3 10 lim x 5 h(x) does not exist.
Infinite limits Look at a graph of the function f (x) = ln(x). The graph has a vertical asymptote at x = 0. As x 0 + the function ln(x) gets more and more negative. We say lim ln(x) =. x 0 +
For every n the line x = π/2 + nπ is a vertical asymptote. We can see that the one sided limits are lim tan(x) = and lim tan(x) =. x π/2 x π/2 +
The line x = a is a vertical asymptote of y = f (x) if at least one of the following is true lim x a f (x) = or lim x a + f (x) = or lim x a f (x) = or
Calculating limits graphically For the function R whose graph is given, state the following 1 lim x 2 R(x) 2 lim x 5 R(x) 3 lim x 3 R(x) 4 lim x 3 + R(x) 5 The equations of the vertical asymptotes.
Calculating limits graphically 1 lim x 2 R(x) = 2 lim x 5 R(x) = 3 lim x 3 R(x) does not exist because lim x 3 R(x) = while lim x 3 + R(x) = 4 The equations of the vertical asymptotes are x = 3, x = 2 and x = 5.
A function f is continuous at a if lim f (x) = f (a). x a This definition requires three things to happen. 1 f (a) exists 2 lim x a + f (x) = f (a) 3 lim x a f (x) = f (a)
Graphs of Continuous Functions This function is continuous at every point except x = 2, 2, 4 and 6.
Continuous Functions Any function defined by the following functions is continuous at every point in its domain. polynomials root functions trigonometric functions inverse trigonometric functions exponential functions and logarithmic functions
Examples of Continuous Functions All of the following functions are continuous wherever they are defined. ( ) f (x) = sin 3+x x 2 +1 g(t) = e 3t/4+ln(2t) r(x) = x 3 +1 2x 5 m(t) = 3t + π tan 1 (4 cos(t π))
Where are the following functions continuous 1 x 2 sin(x) (x 3)(x+4) is continuous except when x = 3 and x = 4 2 ln(x 2 1) is continuous except when 1 x 1 3 e x/(x 1) is continuous except when x = 1.
Evaluating Limits of Continuous Functions This is the easiest thing we will do all class all quarter. If f (x) is continuous at a then lim f (x) = f (a). x a
Evaluating Limits of Continuous Functions Find the following limits. 1 lim x 3 3πx+1 x 2 6 = 2 lim y 2 sin 3y+10 5y 2 3 3 lim x 2 3cx+12 7cx+1 = 9π+1 3 = sin(2) 17 12 6cx 1 14c for any c < 2 and c 1/14. When evaluating limits in this course evaluating the function at the point should always be your first approach. If this fails then try pugging in numbers close to a. After you make a guess you must justify your answer.
Find lim h 0 (2 + h) 3 8. h Plugging in h = 0 we get 0 0. (2 + h) 3 8 lim h 0 h = lim h 0 (2 + h) 3 8 h 8 + 12h + 6h 2 + h 3 8 = lim h 0 h 12h + 6h 2 + h 3 = lim h 0 h = lim 12 + 6h + h 2 h 0 = 12. If two functions agree everywhere except at one point a then the limits approaching a are the same.
Find lim t 0 1/(7 + 3t) 1/7. t Plugging in t = 0 we get 0 0. 1/(7 + 3t) 1/7 lim t 0 t 7 (7 + 3t) = lim t 0 t(7 + 3t)(7) 3t = lim t 0 t(7 + 3t)7 3 = lim t 0 (7 + 3t)7 = 3 49.
Rationalize the denominator Find lim t 16 Plugging in t = 16 we get 0 0. 16 t 4 t. lim t 16 16 t 4 t (16 t)(4 + t) = lim t 16 (4 t)(4 + t) (16 t)(4 + t) = lim t 16 (16 t) = lim t 16 4 + t = 8
Find ( 3 lim x 0 x 3 ) x 3. + x Plugging in x = 0 we get. ( 3 lim x 0 x 3 ) x 3 + x = lim x 0 3(x 2 + 1) 3 x 3 + x 3x 2 = lim x 0 x 3 + x = lim x 0 = 3 0 0 2 + 1 = 0. 3x x 2 + 1
Find Plugging in x = 3 we get 0 0. x 3 27 lim x 3 x 3 x 3 27 lim x 3 x 3 = lim x 0 (x 3)(x 2 + 3x + 9) x 3 = lim x 0 x 2 + 3x + 9 = 27.
Limit Laws Suppose c is a constant and the limits exist. Then lim f (x) and lim g(x) x a x a 1 lim x a [ f (x) + g(x) ] = limx a f (x) + lim x a g(x) 2 lim x a [ f (x) g(x) ] = limx a f (x) lim x a g(x) 3 lim x a [ cf (x) ] = c limx a f (x) 4 lim x a [ f (x) g(x) ] = limx a f (x) lim x a g(x) 5 if lim x a g(x) 0 f (x) lim x a g(x) = lim x a f (x) lim x a g(x)
Find ( lim cos( 3π ) t 1 t + 1 ) 3 (3(t + 1) 2 4t). This looks like a mess. Let s look at the graph.
Find ( lim cos( 3π ) t 1 t + 1 ) 3 (3(t + 1) 2 4t). This looks like a mess. Let s look at the graph.
Here is the graph of the function ( g(t) = cos( 3π ) t + 1 ) 3 (3(t + 1) 2 4t) along with f (t) = 2(3(t + 1) 2 4t) and h(t) = 4(3(t + 1) 2 4t). Notice that as t 1 all three functions are approaching 0.
As these are continuous functions and defined at t = 1 we get and lim t 1 2(3(t + 1)2 4t) = 2(3( 1 + 1) 4( 1) = 0 lim 4(3(t + 1)2 4t) = 4(3( 1 + 1) 4( 1) = 0. t 1 The function that we are interested in ( g(t) = cos( 3π ) t + 1 ) 3 (3(t + 1) 2 4t) is sandwiched in between so lim t 1 g(t) must be zero as well.
Sandwich Theorem This theorem makes the previous slide precise. Theorem If f (x) g(x) h(x) in some interval around a and lim (f (x)) = lim (h(x)) = L x a x a then lim (g(x)) = L x a
Find lim x 14 x 14 14 + x. First we plug in x = 14 and get 14 14 14 + 14 = 0 28 = 0. As this is a continuous function defined at 14 we have lim x 14 x 14 14 + x = 0.
Limit Laws and Graphs Here is the graph of a function f (x) Find the following limits. 1 lim x 2 (f (x) 2 ) 2 lim x 2 (f (x)f (x + 3)) 3 lim x 5 f (x)(2 f (x 3)) 2
Let f (x) = x 1 (x+2)x 2. 1 Find all vertical asymptotes of f (x). 2 Find the one sided limits of f (x) as x approaches the asymptotes. The function is a rational function so it is continuous at every point where it is defined. It is defined everywhere the denominator is not zero, which is when x = 0 and x = 2. It is important to keep signs right. We must be careful about the terms close to zero.
As x 0 +, x 2 approaches 0 from the right. Also when x 0, x 2 approaches 0 from the right. As x 0 the numerator x 1 approaches 1 and x + 2 approaches 2. Thus as x 0 from either side the numerator is negative and the denominator is small and positive. The fraction is thus large and negative. Because of this lim x 0 x 1 (x + 2)x 2 =.
As x 2 + the numerator x 1 approaches 3 and the denominator approaches 0 from the positive side. As the numerator is negative and the denominator is small and positive the fraction is large and negative. Because of this x 1 lim x 2 + (x + 2)x 2 =. As x 2 the numerator x 1 approaches 3 and the denominator approaches 0 from the negative side. As the numerator is negative and the denominator is small and negative the fraction is large and positive. Thus x 1 lim x 2 (x + 2)x 2 =.
When we graph the function we can see that there are vertical asymptotes at x = 2 and x = 0.
1 Find all vertical asymptotes of x 1 (x+2)x 2. 2 Find the one sided limits of f (x) as x approaches the asymptotes. The function is a rational function so it is continuous at every point where it is defined. It is defined everywhere the denominator is not zero, which is when x = 0 and x = 2. It is important to keep signs right. We must be careful about the terms close to zero.
As x 0 +, x 2 approaches 0 from the right. Also when x 0, x 2 approaches 0 from the right. As x 0 the numerator x 1 approaches 1 and x + 2 approaches 2. Thus as x 0 from either side the numerator is negative and the denominator is small and positive. The fraction is thus large and negative. Because of this lim x 0 x 1 (x + 2)x 2 =.
As x 2 + the numerator x 1 approaches 3 and the denominator approaches 0 from the positive side. As the numerator is negative and the denominator is small and positive the fraction is large and negative. Because of this x 1 lim x 2 + (x + 2)x 2 =. As x 2 the numerator x 1 approaches 3 and the denominator approaches 0 from the negative side. As the numerator is negative and the denominator is small and negative the fraction is large and positive. Thus x 1 lim x 2 (x + 2)x 2 =.
Limit Laws Suppose c is a constant and the limits exist. Then lim f (x) and lim g(x) x a x a 1 lim x a [ f (x) + g(x) ] = limx a f (x) + lim x a g(x) 2 lim x a [ f (x) g(x) ] = limx a f (x) lim x a g(x) 3 lim x a [ cf (x) ] = c limx a f (x) 4 lim x a [ f (x) g(x) ] = limx a f (x) lim x a g(x) f (x) limx a f (x) 5 lim x a g(x) = lim x a g(x) if lim x a g(x) 0
To infinity and beyond
Definition of Limits at We write lim f (x) = L x when the values of f (x) can be made arbitrarily close to L by taking x sufficiently large. Similarly we write lim f (x) = L x when the values of f (x) can be made arbitrarily close to L by taking sufficiently large negative values for x.
Limits at Graphically Consider the function f (x) = 3 2e x As x gets large f (x) gets closer to 3. The graph of f (x) gets closer to the graph of y = 3 and we write lim 3 x 2e x = 3.
Limits at Graphically As x gets more and more negative 3 2e x gets more and more negative as well. We write lim 3 x 2e x =.
Horizontal Asymptotes The line y = L is a horizontal asymptote of the curve y = f (x) if either lim f (x) = L or lim f (x) = L. x x Thus y = π/2 and y = π/2 are horizontal asymptotes and lim x tan 1 (x) = π/2 and lim tan 1 (x) = π/2 x
Limits of polynomials at If r > 0 then lim x x r = If r < 0 then lim x x r = 0 If r is even then lim x x r = If r is odd then lim x x r =
Limits Laws at Suppose c is a constant and the limits exist. Then lim f (x) and lim g(x) x x 1 lim x [ f (x) + g(x) ] = limx f (x) + lim x g(x) 2 lim x [ f (x) g(x) ] = limx f (x) lim x g(x) 3 lim x [ cf (x) ] = c limx f (x) 4 lim x [ f (x) g(x) ] = limx f (x) lim x g(x) f (x) limx f (x) 5 lim x g(x) = lim x g(x) if lim x g(x) 0
Extended Real Numbers When working with limits it is useful to use the extended real numbers. = + = + a = a = if a > 0 a = if a < 0 =???
Rational Functions Find the following limits. lim x ( 5x 3 + 2(r 1)x) x lim 3/2 +2πx 7 2x x 1 5x lim 3 +2(r 1)x x x 4 +1
Polynomials With polynomials the largest term dominates. To see this we factor out the largest power of x. lim ( 5x 3 + 2(r 1)x) = lim x 3 ( 5 + 2(r 1)/x 2 ) x x = lim lim ( 5 + 2(r 1)/x 2 ) x = ( )( 5) = x x 3
Divide by the largest power of the denominator For functions like this we divide top and bottom by the largest power of the denominator. x 3/2 + 2πx 7 2x lim x x 2 1 = lim x (1/x 2 )(x 3/2 + 2πx 7 2x) (1/x 2 )(x 2 1) = lim x x 1/2 + 2πx 5 2/x) 1 1/x 2 = lim x (x 1/2 + 2πx 5 2/x) lim x (1 1/x 2 ) = 0 + + 0 1 0 =
Divide by the largest power of the denominator For this problem we want to divide by x 2 = x 4. 5x 3 + 2(r 1)x lim x x 4 + 1 (1/x 2 )( 5x 3 + 2(r 1)x) = lim x (1/x 2 ) x 4 + 1 = lim x 5x + 2(r 1)/x 1 + 1/x 4 = lim x ( 5x + 2(r 1)/x) lim x 1 + 1/x 4 = 5( ) + 0 1 =
( ) lim x 2 + 3x + 2 x x We rationalize the numerator by multiplying the expression by x 2 + 3x + 2 + x x 2 + 3x + 2 + x ( ) lim x 2 + 3x + 2 x x ( = lim x 2 + 3x + 2 x x x 2 + 3x + 2 x 2 = lim x x 2 + 3x + 2 + x 3x + 2 = lim x x 2 + 3x + 2 + x ) x 2 + 3x + 2 + x x 2 + 3x + 2 + x
Then we multiply numerator and denominator by 1/x. ( ) lim x 2 + 3x + 2 x x = lim x (3x + 2)(1/x) ( x 2 + 3x + 2 + x)(1/x) 3 + 2/x = lim x ( 1 + 3/x + 2/x 2 + 1) lim x (3 + 2/x) = lim x ( 1 + 3/x + 2/x 2 + 1) 3 = 1 + 1 = 3 2
For all real numbers a > 0, b and c find ( ) ax 2 + bx + c 5x and lim x lim x ( ) ax 2 + bx + c 5x As in the previous problem we will multiply the expression times ax 2 + bx + c + 5x ax 2 + bx + c + 5x and then factor out the largest power of the denominator.
( ) lim ax 2 + bx + c 5x x ax = lim ( ax 2 + bx + c 5x) 2 + bx + c + 5x x ax 2 + bx + c + 5x ax 2 + bx + c 25x 2 = lim x ax 2 + bx + c + 5x = lim x (a 25)x 2 + bx + c ax 2 + bx + c + 5x (1/x)(a 25)x 2 + bx + c) = lim x (1/x)( ax 2 + bx + c + 5x) (a 25)x + b + c/x = lim x a + b/x + c/x 2 + 5 = lim x ((a 25)x + b + c/x) lim x ( a + b/x + c/x 2 + 5) = lim x ((a 25)x + b + 0) a
( ) lim ax 2 + bx + c 5x = lim x ((a 25)x + b + 0) x a If a > 25 then the leading term has a positive coefficient and the limit approaches. If a < 25 then the leading term has a negative coefficient and the limit approaches. If a = 25 then the limit becomes b/5. The limits as x are found the the same way.
Find lim x ( ) lim 4x 4 + 5x 3 + 10 (2x 2 + x). x Again we will multiply by the conjugate and divide by the highest power of x in the denominator. ( ) 4x 4 + 5x 3 + 10 (2x 2 + x) ( = lim 4x 4 + 5x 3 + 10 (2x 2 + x) x = lim x (4x 4 + 5x 3 + 10) (2x 2 + x) 2 4x 4 + 5x 3 + 10 + (2x 2 + x) = lim x (4x 4 + 5x 3 + 10) (4x 4 + 4x 3 + x 2 ) 4x 4 + 5x 3 + 10 + (2x 2 + x) = lim x x 3 x 2 + 10 4x 4 + 5x 3 + 10 + (2x 2 + x) ) 4x 4 + 5x 3 + 10 + (2x 2 + 4x 4 + 5x 3 + 10 + (2x 2 +
x 1 + 10/x 2 = lim 4 + 5/x + 10/x x 2 4 + 2 + 1/x 2 =
1 Find all vertical asymptotes of x 1 (x+2)x 2. 2 Find all horizontal asymptotes of x 1 (x+2)x 2. 3 Find the one sided limits of f (x) as x approaches the asymptotes. The function is a rational function so it is continuous at every point where it is defined. It is defined everywhere the denominator is not zero, which is when x = 0 and x = 2. It is important to keep signs right. We must be careful about the terms close to zero.
As x 0 +, x 2 approaches 0 from the right. Also when x 0, x 2 approaches 0 from the right. As x 0 the numerator x 1 approaches 1 and x + 2 approaches 2. Thus as x 0 from either side the numerator is negative and the denominator is small and positive. The fraction is thus large and negative. Because of this lim x 0 x 1 (x + 2)x 2 =.
As x 2 + the numerator x 1 approaches 3 and the denominator approaches 0 from the positive side. As the numerator is negative and the denominator is small and positive the fraction is large and negative. Because of this x 1 lim x 2 + (x + 2)x 2 =. As x 2 the numerator x 1 approaches 3 and the denominator approaches 0 from the negative side. As the numerator is negative and the denominator is small and negative the fraction is large and positive. Thus x 1 lim x 2 (x + 2)x 2 =.
Sketch a graph of a function f (x) with the following properties. 1 has horizontal asymptotes y = 0 and y = 4 2 has vertical asymptotes x = 3 and x = 4 3 lim x 4 f (x) = 4 lim x 3 f (x) does not exist.
Graphical limits Composition of functions. Change of variables.