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LIMIT MENU

LIMIT MENU. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006. Basic ideas and exercises ( Slides 2-6 ) Limits using a Calculator ( Slides 7-8 ) “-  definition”, limit validation & properties” ( Slides 9-13 )

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LIMIT MENU

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  1. LIMIT MENU A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 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)

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

  3. GEOMETRIC INTERPRETATION OF THE LIMIT Let’s see now the geometric interpretation of lim (x2 – 4)/(x-2) . x  2 The graph f(x) is the line y=x+2 with a hole at x=2. See the figure on right y=x+2 Y 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. hole 4 º y = f(x) is a line with a hole at (2,4). Other way of saying this is 1) when x2- ( from the left of 2) 2) the value of f(x) 4 (see figure) 2 X 2) when x2+ (from the right of 2) . the value of f(x) 4 (see figure) -2 2 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 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 Ex.1 Find lim. x  -10 Replacing x= -10 on (x2 +x+1) / (x+1) we get (100-10+1) / ( -10+1)= - 91/9 Replacing x= 1/2 on (2x2 - x) / (2x+1) we get 0/2 = 0. Ex.2 Find lim . x 1/2 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.

  5. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 Ex.4 Find lim . x- 2 Replacing x = - 2, we got 0/0. So we factor out x+2 lim x- 2 = lim . 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0 = lim . x0

  6. SUMMARY OF FINDING LIMITS A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 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.

  7. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 USING A CALCULATOR TO GUESS THE LIMIT 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) 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 Example: Guess the value of lim . x0 Graph y = sin x / x for -1< x <1 & Scale .001 . -1< y <2 & Scale .0001 Y 2 ( 0.03 , 0.99985) (- 0.03 , 0.99985) o 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). f(x)=sin x / x X - 1 1

  8. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 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 ±1/10nfor n=1,2,3,4,5,… lim x0 Example 1: Guess the value of For the values of x near to 0, both sin x and x has the same sign; so we try only the positive numbers. x So, it seems that lim =1 . x0 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.998334166468 0.999983333417 0.999999833333 0.999999998333 0.999999999983 0.999999999999 (Proof will come later) X Example 2: Find L, if L=lim . x  0 0.0001 0.000010.000001 0.0000001 0.00000001 0.5000250… 0.5000025… 0.5000002… 0.5000000… 0.5000000… So, L=0.5

  9. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 RIGOROUS DEFINITION OF LIMIT The limit of f(x), when x0, is L means that “ f(x) can take values as close as we want to L as long as 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| < ” Y f(x)=2x+3 Example 1:Prove that lim (2x+3) = - 1. x -2 Let >0be 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+). 3 See figure on the right (-2- ,-2+ ) X So, we need to find a  >0 such that satisfies:| x+2 | <   | 2x+3- (-1)| < -2 - 1.5 -1+-1 -1+ But | 2x +3- (-1)|= |2x+4| = 2|x+2| < 2 So we have to restrain 2 to be less than .e.gchoose < ½ and itwill do the work. This proves that lim (2x+3) = -1. x - 2

  10. Example 2. Prove that lim (x2 +1) = 5. x  2 A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 Given any  > 0 we look for a  >0 , such that for any x in (2- , 2+ ) the values of f(x) remain in (5- , 5+ ). See figure below. So, we need to find a  >0 that satisfies: . | x-2 | <   | (x2 +1) - 5| < f(x) = x2 +1 Y Since |x2 - 4| = |x+2| | x-2| < | x+2|  Let us assume that  < 1 and then it allows us to find a bound for |x+2 | 5 + 5 5 - From |x-2| <  < 1, we can write • 1 < x-2 < 1 -1 < x-2 < 1 4 + 4 + 4 , so - 5 < 3< x+2 < 5 | x +2 |< 5 So |x2 - 4 | = |x+2| |x-2| < 5 . Now imposing that 5<  , we get < /5 X So if  < 1 pick < /5 & if ≥ 1 pick  =1 2 So = min{ / 5 , 1} and itwill do thework. (2 -  , 2+ ) This proves that lim (x2 +1) = 5. x  2

  11. ADDITIONAL PROPERTIES OF LIMIT A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 Theorem: Given two functions A & B If lim A = L. . . x c & lim B = M, then: . x c • lim (A ± B) • x  c = L ± M 2)lim (A•B). . x c. = L• M 3) lim An. . x c = Ln 4) lim (A/B) . x c = L/ M when M ≠ 0 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. We’ll accept this theorem without proof. Examples: 1) lim 3x x 7 = lim 3 • . x 7 lim x = 3•7 = 21x 7 2) lim x2. x - 2 = lim x • lim x. . x- 2 x-2 = -2 •(-2)=(-2)2 = 4 3) lim (5x2-2x-28) = .x 3 5(3)2 – 2• 3 –28 = 45-6-28 = 11.

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

  13. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 PIECEWISE - DEFINED FUNCTIONS (x2+2x)/ x if x<1 Example: Let f be defined byf(x) = x2 - 6x+8 If 1≤ x< 5 4 If x > 5 Find each limit (if the limit doesn’t exist, write DNE) 1) lim f(x) =. x -2 f(-2)= (4 - 4) /-2 = 0 2) lim f(x) =. x 0 Since f(0)=0/0, we try lim (x2+2x)/ x = lim (x+2) =2 . . x 0 x 0 So lim f(x) = 2. x 0 Since f(x)= (x2+2x)/ x is defined on the left of x=1 and . f(x)= x2 - 6x+8 is defined on the right of x=1, it’s necessary to calculate both limits when x1- & when x 1+ 3) lim f(x) =. x 1 lim f(x) = lim (x2+2x)/ x = 3x 1 -- x 1-- So lim f(x) = 3. x 1 lim f(x) = lim (x2 - 6x+8) = 3= x 1+ x 1+

  14. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 4) lim f(x) =. x 3 lim (x2 - 6x+8) = x 3 9-18+8 = - 1 5) lim f(x) = lim (x2 - 6x+8) = 0 . x 4x 4 6) lim f(x) . x 5 Since to the left of x=5, f is defined as f(x)= x2 - 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 (x2-6x+8) = 3x 5-- x 5-- So lim f(x) DNE. x 5 lim f(x) = 4 x 5+

  15. GRAPH OF A PIECEWISE - DEFINED FUNCTION A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 (x2+2x)/ x if x<1 & x≠ 0 x2 - 6x+8 If 1≤ x  54 If x > 5 Graph f where f(x)= 1) Graph y= (x2+2x)/ x = x+2 . with holes at x=0 & x=1. y=4 for x>5 Y y=4 4 Constraining the domain of f to x<1. See figure on the right. 3 2) Graph y= x2 - 6x+8 2 Constraining the domain of f to 1≤ x < 5 3) Finally graph y=4 3 X -2 1 5 Constraining the domain of f to x > 5 - 1 4) The graph of f(x) is the . blue one on the right. y= x2- 6x+8 for 1≤ x  5 y= x2- 6x+8 y=x(x-2)/ x for x<1 y=x+2

  16. FINDINGLIMITS FROM THE GRAPH OF A FUNCTION A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 (x2+2x)/ x if x<1 & x≠ 0 Let f be defined byf(x)= x2 - 6x+8 If 1≤ x  5 4 If x > 5 See the graph f(x) on the right. Y Find the following limits: 1) lim f(x) =. x -2 0 limit when x -2– is 0 4 limit when x -2+ is 0 3 2) lim f(x) =. x 1 3 limit when x 1– is 3 2 limit when x 1+ is 3 3) lim f(x) =. x 3 -1 limit when x 3– is -1 X -2 3 limit when x 3+ is -1 1 5 - 1 lim when x 5– lim is 3 4) lim f(x) = . x 5 DNE lim when x 5+ lim is 4

  17. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 MORE ON LIMITS Two theorems will help us to deal with limits when the function involved is continuous or better when it is differentiable. 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 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 Since ln is is a continuous function we can interchange ln and lim . = ln lim [ ] . . x1 Using lim (A+B) = lim A+lim B = ln [ lim + lim 5 ]. x1 x1 Interchanging lim and = ln [ + 5 ]. x1 = ln [ 2+5 ] = ln7

  18. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 Ex 3: lim sin(tan2 ( ) ) =.. x1/2 Interchanging lim & sin =sin ( lim [ tan2 ( )]. x1/2 Interchanging lim &  tan2 = sin[ tan2 [ lim ( )] . x1/2 Evaluating … =sin [ tan2(/3)] = sin [ ( )2 ] = sin (3) = 0 Ex 4: lim = L , find L. x1/2 Applying ln on both sides … ln [ lim ] = ln L . x1/2 Interchanging ln & lim ... lim ln [ ] = ln L x1/2 lim x1/2 = lim.. x1/2 = -5/6 So, ln L = lim = . x1/2 From we get

  19. A Wang & O Schiappacasse Math Dept, Alexandria, NVCC 2006 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.

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

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

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