1. LIMIT MENU Basic ideas and exercises ( Slides 2-6 ) Limits using a Calculator ( Slides 7-8 ) “  -  definition”, limit validation & properties” ( Slides 9-13 ) Piecewise-defined functions: graph and limits ( Slides 14-18 ) L’Hôpital Rule and type of limits 0/0 or ∞/∞ ( Slides 19-21 ) A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

2. will be used to represent the real number to which f(x) approaches when x x c ( with x ≠ c)c) In our example, since x  2, we can rewrite f(x) as So, lim f(x) =. x 22 To introduce the concept of limit we’ll use the function We know that f(2)is undefined (since f(2) = ). In general, to find the limit of f(x) when x x c we need to investigate the behavior of the function f(x) as x gets near to c (where c may or may not be in the domain of f). = 2+2 = 4lim (x+2) x  2 lim f(x) x  c We are interested in assigning a value in according with the behavior ( if there is one) of f(x) as x gets near 2. So, the behavior of f is being the line f(x)=x+2 not containing the point (2, 4). LIMITS A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

3. The graph f(x) is the line y=x+2 with a hole at x=2. See the figure on right Other way of saying this is 1) when x  2 - ( from the left of 2) the value of f(x)  4 (see figure) Let’s see now the geometric interpretation of lim (x 2 – 4)/(x-2). x  2 y = f(x) is a line with a hole at (2,4). -2 2 2 4 X Y hole º GEOMETRIC INTERPRETATION OF THE LIMIT The value of lim f(x) when x  2 following the behavior of f(x) (which is the line y=x+2 ) So the limit is f(2) = 4. y=x+2 2) when x  2 + (from the right of 2). the value of f(x)  4 (see figure) We see that the limit from the left = 4 & the limit from the right = 4, so the limit exists and equals 4.

4. CALCULATING BASIC LIMITS Ex.1 Find lim. x  -10 Replacing x= -10 on (x 2 +x+1) / (x+1) we get (100-10+1) / ( -10+1)= - 91/9 Ex.2 Find lim. x  1/2 Replacing x= 1/2 on (2x 2 - x) / (2x+1) we get 0/2 = 0.0. Ex.3 Find lim.. x  3 Replacing x=3 we get 0/0 (undefined !) The Remainder Theorem assures us that both polynomials in numerator and denominator have a factor (x-3). Factoring … So lim. x  3 So, the limit does not exist. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

5. Ex.4 Find lim. x  - 2 Replacing x = - 2, we got 0/0. So we factor out x+2 lim x  - 2 Ex.5 Find lim. 3. x  0 Replacing x=0, we get 0/0. Rationalizing the numerator we have x on the numerator : lim x  0 = lim.. x  0 = lim. x  0 = lim. x  - 2 = lim.. x  0 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

6. SUMMARY OF FINDING LIMITS I)If p(x) is a polynomial function, then perform the direct substitution, the value p(a) is the limit: lim p(x) = p(a). x  a II) If f(x) = p(x)/q(x) is a rational function, a quotient of two polynomials p(x) and q(x). First find the values of p(a) and q(a). Then a) If g(a) ≠0, lim f(x) = p(a) / p(a).. x  a b) If g(a) =0 and f(a)≠0, lim f(x) does not exist. x  a c) If both p(a) and g( a) are zero, then we know that x-a is a common factor. Simplify until we get a non zero value for either one of then and repeat procedures (1) and (2). III)If f(x) is a piece-wise defined function, check the limit as x gets near to a from either sides of a, if lim f(x) = lim f(x). x  a - x  a + then the limit exists. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

7. USING A CALCULATOR TO GUESS THE LIMIT Example: Guess the value of lim. x  0 - 1 1 ( 0.03, (- 0.03, Using ZOOM IN you’ll see that when x is close to 0, the value of y is close to 1. So, we guess that the limit goes to 1 (but need to be proved). Y X 0.99985) 2 f(x)=sin x / x o Graph y = sin x / x for -1< x <1 & Scale.001. -1 < y < 2 & Scale.0001 a) Graph the function in a neighborhood of x=c. Use TRACE to move the cursor along the graph ( for values of x near to c) to guess the limit value When the previous method doesn’t work we can use a Graphing Calculator to help us to guess the limit. The following two options could help. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

8. b) Making a table y versus x for values very close to x=c (from the left and from the right). For instance take x= c ± 1/10 n for n=1,2,3,4,5,… 0.998334166468 0.999983333417 0.999999833333 0.999999998333 0.999999999983 0.999999999999 x So, it seems that lim =1. x  0 For the values of x near to 0, both sin x and x has the same sign; so we try only the positive numbers. Example 1: Guess the value of (Proof will come later) Example 2: Find L, if L= lim. x  0 X 0.0001 0.00001 0.000001 0.0000001 0.00000001 So, L=0.5 lim x  0 0.5000250… 0.5000025… 0.5000002… 0.5000000… 0.5000000… 0.1 0.01 0.001 0.0001 0.00001 0.000001 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

9. The limit of f(x), when x0, x0, is L means that “ f(x) can take values as close as we want to L as long as x x 0”, this can be symbolically expressed as: for any given  >0 we can find a corresponding  >0 for which |x-c| <  implies | f(x)- L| <  ” Example 1:Prove that lim (2x+3) = - 1. x  -2 Let  > 0 be given, we look for a  > 0, such that for any x in (-2- , -2+  ), an open interval containing -2, the values of f(x) remain in (-1- , -1+  ). See figure on the right So, we need to find a  >0 such that satisfies: | x+2 | <   | 2x +3- (-1)| <  So we have to restrain 2  to be less than . e.g choose  < ½  and it will do the work. X Y f(x)=2x+3 -2 3 - 1.5 -1+  -1 - 1+  (-2- ,-2+  ) This proves that lim (2x+3) = -1. x  - 2 RIGOROUS DEFINITION OF LIMIT But | 2x +3- (-1)|= |2x+4| = 2|x+2| < 2  A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

10. 5 5 +  5 -  f(x) = x 2 +1 Example 2. Prove that lim (x 2 +1) = 5. x  2 Given any  > 0 we look for a  >0, such that for any x in (2- (2-, 2+ 2+ ) the values of f(x) remain in (5- (5-, 5+ 5+ ).). So, we need to find a  >0 that satisfies:. | x-2 | <   | (x 2 +1) - 5| <  See figure below. Y 2 (2 - , 2 + )) X Since |x 2 - 4| = |x+2| | x-2| < | x+2|  Let us assume that  < 1 and then it allows us to find a bound for |x+2 | From |x-2| <  < 1, we can write - 1 < x-2 < 1  -1 < < 1 3 < x+2 < 5  | x +2 |< 5 So |x 2 - 4 | = |x+2| |x-2| < 5 . So if  < 1 pick  <  / 5 & if  ≥ 1 pick  =1 - 5 < So  = min{  / 5, 1} and it will do the work. 4 + 4 + 4, so Now imposing that 5  < , we get  <  / 5 This proves that lim (x 2 +1) = 5. x  2 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

11. If lim A = L... x x c = lim 3. x  7 = lim x lim x.. x  - 2 x  -2 3) lim (5x 2 -2x-28) =. x  3 1)lim (A ± B) x  c & lim B = M, then:. x  c 3) lim A n.. x  c 2) lim (AB).. x  c. 4) lim (A/B). x  c Examples: 1) lim 3x x  7 2) lim x 2. x  - 2 5(3) 2 – 2 3 –28 = 45-6-28 = 11. That is, the limit of a sum (difference, product or quotient) of two functions is equal to the sum (difference, product or quotient) of the limits of the functions. ADDITIONAL PROPERTIES OF LIMIT We’ll accept this theorem without proof. lim x = 37 = 21 x  7 = L ± M= L M = L n = L/ M when M ≠ 0 = -2 (-2)=(-2)2 = 4 Theorem: Given two functions A & B A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

12. = lim. x  -1 = lim. x  -1 4) lim. x  -1 lim. x  0 = lim.. x  0 5) lim. x  0 lim. lim. x  0 x  0 ( Hint: ) For x=-1 both denominator and numerator are 0 Factoring … 6) lim.. x  5 lim x  -1 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

13. PIECEWISE - DEFINED FUNCTIONS Example: Let f be defined by f(x) = (x 2 +2x)/ x if x<1 x 2 - 6x+8 If 1≤ x< 5 4 If x > 5 1) lim f(x) =. x  -2 3) lim f(x) =. x  1 f(-2) = (4 - 4) /-2 = 0 Since f(0)=0/0, we try lim (x 2 +2x)/ x = lim (x+2) =2.. x  0 x  0 Find each limit (if the limit doesn’t exist, write DNE) So lim f(x) = 2. x  0 Since f(x)= (x 2 +2x)/ x is defined on the left of x=1 and. f(x)= x 2 - 6x+8 is defined on the right of x=1, it’s necessary to calculate both limits when x  1 - & when x  1 + lim f(x) = lim (x 2 +2x)/ x = 3 x  1 -- x  1 -- lim f(x) = lim (x 2 - 6x+8) = 3 = x  1 + x  1 + So lim f(x) = 3. x  1 2) lim f(x) =. x  0 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

14. 4) lim f(x) =. x  3 6) lim f(x). x  5 5) lim f(x) = lim (x 2 - 6x+8) = 0. x  4 x  4 lim (x 2 - 6x+8) = x  3 9-18+8 = - 1 Since to the left of x=5, f is defined as f(x)= x 2 - 6x+8. & to the right of x=5, f is defined as f(x)= 4, it’s necessary to calculate both limits lim f(x) = lim (x 2 -6x+8) = 3 x  5 -- x  5 -- lim f(x) = 4 x  5 + So lim f(x) DNE. x  5 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

15. X Y 1 y=x+2 -2 2 y= x 2 - 6x+8 3 5 - 1 3 y=x(x-2)/ x for x<1 Graph f where f(x) = 1) Graph y= (x 2 +2x)/ x = x+2. with holes at x=0 & x=1. Constraining the domain of f to x<1. See figure on the right. 2) Graph y= x 2 - 6x+8 Constraining the domain of f to 1≤ x < 5 y= x 2 - 6x+8 for 1≤ x  5 3) Finally graph y=4 Constraining the domain of f to x > 5 4 ( x 2 +2x)/ x if x 5 GRAPH OF A PIECEWISE - DEFINED FUNCTION y=4 y=4 for x>5 4) The graph of f(x) is the. blue one on the right. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

16. FINDING LIMITS FROM THE GRAPH OF A FUNCTION Let f be defined by f(x) = (x 2 +2x)/ x if x<1 & x ≠ 0 x 2 - 6x+8 If 1≤ x  5 4 If x > 5 1) lim f(x) =. x  -2 2) lim f(x) =. x  1 4) lim f(x) =. x  5 Find the following limits: X Y 1 -2 2 3 5 - 1 3 4 See the graph f(x) on the right. 3) lim f(x) =. x  3 lim when x  5 – lim is 3 0 limit when x  - 2 – is 0 limit when x  1 – is 3 limit when x  3 – is -1 limit when x  - 2 + is 0 limit when x  1 + is 3 lim when x  5 + lim is 4 limit when x  3 + is -1 3 DNE A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

17. MORE ON LIMITS Theorem 1: If g is continuous at a and f is continuous at g(a) then lim f( g(x) ) = f( lim (g(x) ). x  a x  a Two theorems will help us to deal with limits when the function involved is continuous or better when it is differentiable. This theorem allows us to interchange f with the lim Ex 1: lim sin( ) =. x  1 sin[ lim ]. x  1 = sin( ) = -(½)  2 Ex 2: lim ln [., ]. x  1 = ln lim [ ].. x  1 = ln [ lim + lim 5 ]. x  1 x  1 = ln [ + 5 ]. x  1 = ln [ 2+5 ]= ln7 Since ln is is a continuous function we can interchange ln and lim. Using lim (A+B) = lim A+lim B Interchanging lim and A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

18. Ex 3: lim sin(  tan 2 ( ) ) =.. x  1/2 =sin ( lim [  tan 2 ( )]. x  1/2 = sin[  tan 2 [ lim ( )]. x  1/2 = sin [  tan 2 (  /3)]= sin [  ( ) 2 ] = sin (3  ) = 0 Interchanging lim & sin Interchanging lim &  tan 2 Evaluating … Ex 4: lim = L, find L. x  1/2 Applying ln on both sides … Interchanging ln & lim... ln [ lim ] = ln L. x  1/2 lim ln [ ] = ln L x  1/2 lim x  1/2 From we get = lim.. x  1/2 So, ln L = lim =. x  1/2 = -5/6 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

19. Theorem (L’Hôpital Rule) : If f & g are differentiable functions such that f(a)=0 & g(a)=0, then Remark: The Theorem is also true if f(a)= g(a)= , meaning a can be any real number or ± . Proof: Check first that Dividing both numerator & denominator by (x-a) and then taking limit on both sides, we get: This proves the theorem. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

20. Ex 1 : 1/1 = 1 Ex 2: Ex 3 : Ex 4: A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006

21. Ex 5: lim.. x   Let y= and apply ln to both sides.. ln y= ln lim. x   = lim.. x   = lim. x   = lim.. x  (Type ∞/∞) Using L’Hopital Rule A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006