1. Precise definition of limits The phrases “x is close to a” and “f(x) gets closer and closer to L” are vague. since f(x) can be arbitrarily close to 5 as long as x approaches 3 sufficiently. How close to 3 does x have to be so that f(x) differs from 5 by less than 0.1? Solving the inequality |(2x-1)-5|<0.1, we get |x-3|<0.05, i.e., we find a number  =0.05 such that whenever |x-3|<  we have |f(x)-5|<0.1

2.  definition of a limit If we change the number 0.1 to other smaller numbers, we can find other  s. Changing 0.1 to any positive real number , we have the following Definition: We say that the limit of f(x) as x approaches a is L, and we write if for any number  >0 there is a number  >0 such that

3. Remark  expresses “arbitrarily” and  expresses “sufficiently” Generally  depends on  To prove a limit, finding  is the key point means that for every  >0 (no matter how small  is) we can find  >0 such that if x lies in the open interval (a- ,a+  ) and x  a then f(x) lies in the open interval (L- ,L+  ).

4. Example Ex. Prove that Sol. We solve the question in two steps. 1. Preliminary analysis of the problem (deriving a value for  ). Let  be a given positive number, we want to find a number  such that But |(4x-5)-7|=|4x-12|=4|x-3|, therefore we want

5. Example (cont.) This suggests that we should choose  =  / . 2. Proof (showing the above  works). Given  choose  If 0<|x-3|< , then |(4x-5)-7|=|4x-12|=4|x-3|<4  Thus Therefore, by definition we have

6. Example Ex. Prove that Sol. 1. Deriving a value for . Let  >0 be given, we want to find a number  such that Since |(x 2 -x+2)-4|=|x-2||x+1|,if we can find a positive constant C such that |x+1|<C, then |x-2||x+1|<C|x-2| and we can make C|x-2|<  by taking |x-2|<  /C  As we are only interested in values of x that are close to 2,

7. Example (cont.) it is reasonable to assume |x-2|<1. Then 1<x<3, so 2<x+1<4, and |x+1|<4. Thus we can choose C=4 for the constant. But note that we have two restrictions on |x-2|, namely, |x-2|<1 and |x-2|<  /C=  /4. To make sure both of the two inequalities are satisfied, we take  to be the smaller of 1 and  /4. The notation for this is  =min{1,  /4}. 2. Showing above  works. Given  >0, let  =min{1,  /4}.

8. Example (cont.) If 0<|x-2|< , then |x-2|<1 ) 1<x<3 ) |x+1|<4. We also have |x-2|<  /4, so |(x 2 -x+2)-4|=|x-2||x+1|<  /4 ¢ 4=  This shows that  can be found by solving the inequality, but no need to solve the inequality: is not unique, finding one is enough

9. Example Ex. Prove that Sol. For any given  >0, we want to find a number  >0 such that By rationalization of numerator, If we first restrict x to |x-4|<1, then 3<x<5 and

10. Example (cont.) Now we have and we can make by taking Therefore If  >0 is given, let When 0<|x-4|<  we have firstly and then This completes the proof.

11. Proof of uniqueness of limits (uniqueness) If and then K=L. Proof. Let  >0 be given, there is a number  1 >0 such that |f(x)-K|<  whenever 0<|x-a|<  1. On the other hand, there is a number  2 >0 such that |f(x)-L|<  whenever 0<|x-a|<  2. Now put  =min{  1,  2 } and x 0 =a+  Then  |f(x 0 )-K|<  and |f(x 0 )-L|< . Thus |K-L|=|(f(x 0 )-K)-(f(x 0 )-L)| · |f(x 0 )-K|+ |f(x 0 )-L|<2 . Since  is arbitrary, |K-L|<2  implies K=L.

12.  definition of one-sided limits Definition: If for any number  >0 there is a number  >0 such that then Definition: If for any number  >0 there is a number  >0 such that then

13. Useful notations 9 means “there exist”, 8 means “for any”.  definition using notations such that there holds

14. M-  definition of infinite limits Definition. means that 8 M>0, 9  >0, such that whenever  Remark. M represents “arbitrarily large”

15. Negative infinity means

16. Continuity Definition A function f is continuous at a number a if Remark The continuity of f at a requires three things: 1. f(a) is defined 2. The limit exists 3. The limit equals f(a) otherwise, we say f is discontinuous at a.

17. Continuity of essential functions Theorem The following types of functions are continuous at every number in their domains: polynomials algebraic functions power functions trigonometric functions inverse trigonometric functions exponential functions logarithmic functions

18. Example Ex. Find the limits:(a) (b) Sol. (a) (b)

19. Continuous on an interval A function f is continuous on an interval if it is continuous at every number in the interval. If f is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.

20. Continuity of composite functions Theorem If f is continuous at b and then In other words, If g is continuous at a and f is continuous at g(a), then the composite function f(g(x)) is continuous at a.

21. Property of continuous functions The Intermediate Value Theorem If f is continuous on the closed interval [a,b] and let N be any number between f(a) and f(b), where Then there exists a number c in (a,b) such that f(c)=N.

22. Example The intermediate value theorem is often used to locate roots of equations. Ex. Show that there is a root of the equation between 1 and 2. Sol. f(1)=-1 0, there exists a number c such that f(c)=0.

23. Limits at infinity Definition means for every  >0 there exists a number N>0 such that |f(x)-L|<  whenever x>N. means 8  >0, 9 N>0, such that |f(x)-L|<  whenever x<-N.

24. Properties All the properties for the limits as x ! a hold true for the limits as x !1 and Theorem If r>0 is a rational number, then If r>0 is a rational number such that is defined for all x, then

25. Examples Ex. Find the limits (a) (b) Sol. (a) (b)

26. Horizontal asymptote Definition The line y=L is called a horizontal asymptote if either or For instance, x-axis (y=0) is a horizontal asymptote of the hyperbola y=1/x, since The other example, both and are horizontal asymptotes of

27. Infinite limits at infinity Definition means 8 M>0, 9 N>0, such that f(x)>M whenever x>N. means 8 M>0, 9 N>0, such that f(x) N. Similarly, we can define and

28. Homework 3 Section 2.4: 28, 36, 37, 43 Section 2.5: 16, 20, 36, 38, 42 Section 2.6: 24, 32, 43, 53 Page 181: 1, 2, 3, 5, 7