FINDING LIMITS 9/8/2017 CLASS NOTES

Transcription

1 FINDING LIMITS 9/8/2017 CLASS NOTES

2 QUICK REVIEW OF THURSDAY S CLASS Crudely speaking, we can use the limit procedure to find values f(a) of a funcxon f(x) when simply plugging in x=a does not work In calculus limits are essenxal because in most cases they are needed to find the instantaneous rate of change of a func\on at a point. (This is same as the slope of the tangent at that point).

3 Limit of slopes of secant lines As seen before, we can find the slopes of tangents by looking for the value approached by the slope of secants Note that the limit, here the slope of the tangent, may not always exist! To find exact value, we look for a number L such that f(x) can be as close to L as possible.

4 SituaXons where you cannot just plug in a and find f(a) u u Bad cases (singularixes) f(x) goes to infinity graph has verxcal asymptote f(x) oscillates

5 SingulariXes verxcal asymptote 1.0 Where does this graph have verxcal asymptote? Graph of f(x) = 1/x

6 SingulariXes- - oscillaxon Graph of sin(π/x) - - oscillates between - 1 and 1 near zero

7 A good case where limit exists Graph of x sin(π/x) Squeezed between x and x because sin(π/x) is always 0.10 between - 1 and

8 Graph of x 2 sin(π/x) Graph of x 2 sin(π/x) is squeezed between Graph of x 2 and graph of x Why do we know that graph can get as close to 0 as we want, near 0?

9 Squeeze theorem If h(x) f (x) g(x) then Lim x a and Lim x a f (x) = L h(x) = Lim x a g(x) = L In plain English, if f(x) is squeezed between h(x) and g(x) and h and g approach the same value L then f also has to approach L because it has nowhere else to go.

10 Exercise Find the limit of xe sin( pi/x) As x approaches zero from the posixve side.

11 One- sided Limits SomeXmes a funcxon may approach different values as x approaches a from the leg and the right. Example: already saw 1/x approaches from the right of 0 and - from the leg of 0. Here is another example: What is the value of the Floor funcxon as x approaches 0 from the leg? (Recall: Floor(x) = Greatest integer x)

12 Graph of Floor funcxon

13 One- sided Limits (conxnued) We say that Lim x 0 + Floor(x) = 0 And that Lim Floor(x) = 1 x 0

14 Limits, Tangent Lines, Secant Lines The slope of the secant line gives an approximate rate of change (average rate of change) of funcxon from A to B. For example, in previous class we found growth of A(t) from A(t) to A(t+1). The slope of the tangent lines gives the rate of change at the point P instantaneous rate of change. By looking at slopes of secant lines as A and B get closer and to P we can figure out the slope of tangent line, using limit process Note that tangent need not exist! If the curve has a sharp point or a break it won t have a well- defined tangent.

15 Another good case: A simple example of limits Remember that you cannot just plug in x=1. (Why?) Hint: How did we find slope of tangent line?

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17 When x = 1.1, 1.01, 1.001, etc., x+1 is 2.1, 2.01, 2.001, etc and we can see that x+1 can be made as close to 2 as we want by making x close enough to 1.

18 USE THIS: Problem

19 PROBLEM (CONTD) In we cannot cancel t- 1 ager plugging in t=1, because you get 0/0. But as before you can cancel t- 1 when t is close to 1, say like 1.1, 1.01,1.001, etc., Then the value of approaches (1+1)(1 2 +1)/( ) = 4/3. Here we are using limit laws 1,2,4 and 5 (Read!!)

20 Warning about just using numbers Ager class I was asked if it was okay to find the limit by plugging in various values, for example by plugging in 1, 1.1, 1.01, etc., in previous problem. This would certainly help but it is not enough! Example that shows why it is not enough: Sin(π/x) = 0 for x = 1, x= ½ = 0.5, x = 1/3 = x = ¼ = 0.25, etc., But sin(π/x) does not approach 0 as x - > 0. In fact as we saw it keeps oscillaxng between 1 and - 1

21 Precise definixon of limit (2.4) What do you mean We can make x+1 as close to 2 as possible by choosing x close enough to 1?

22 Precise definixon of limit (cont.d) It means distance between x+1 and 2 can be made as small as possible by choosing x close enough to 1. i.e, by making distance between x and 1 small enough. To write distance is as small as possible symbolically we use the Greek lewers ε and δ. We say (x+1)- 2 < ε for any ε however small If we choose δ such that x- 1 < δ. In pracxce this means solving for δ in terms of ε.

23 Precise definixon of limit (cont.d) We say (x+1)- 2 < ε for any ε however small If we choose δ such that x- 1 < δ. In pracxce this means solving for δ in terms of ε. But in this case x+1-2 is just x- 1. So in this case ε = δ. In otherwords, if we want x+1 to be , we can choose x =

24 Precise definixon of limit (cont.d) Problem Prove the following using the precise definixon of limits (i.e, the ε and δ definixon of limit).

25 Precise definixon of limit (cont.d) Problem

26 Precise definixon of limit (cont.d) Problem

27 Precise definixon of limit Problem (cont.d) So basically we are trying to solve for ε in terms of δ. Here, for any ε however small, by choosing δ = ε/2, we get the desired result. In pracxce, ε and δ method basically involves y- L in terms of x- a where we are trying to show that y = f(x) is approaches the limit L as x approaches a.

28 A pracxce problem: Lim h h 3 h =?

29 A pracxce problem: (cont.d)

30 A pracxce problem: (cont.d)