f ( x) = L ( the limit of f(x), as x approaches a,

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1 Math 1205 Calculus Sec. 2.4: The Definition of imit I. Review A. Informal Definition of imit 1. Def n : et f(x) be defined on an open interval about a except possibly at a itself. If f(x) gets arbitrarily close to (as close to as we like) for all x sufficiently close to a, we say that f approaches the limit as x approaches a and we write: lim equals ). f ( x) = ( the limit of f(x), as x approaches a, x!a 2. Note: (1) x! a means that you approach x=a from both sides of a. (2) f(a) does not have to be defined. B. Solving an inequality 1. Fill in the blanks: To say that x - 3 < 1 means that x is less than units from. 2. Solve x - 3 < 1 II. Introduction A. Consider the function f(x)=-2x+5. How close to a=1, must we hold x to be sure that f(x) lies within 1.5 units of f(a)=3? -ε x 1 a x 2 How do we determine the value of delta,δ? Method 1: Set f(x)= and set f(x)=-ε and solve for x 1 = a-δ and x 2 = a+δ and then determine δ

2 Method 2: What we want to know is: When is f(x)- <1.5? f(x)-3 < 1.5! (-2x+5)-3 < 1.5! -2x+2 < 1.5! -1.5 < -2x+2 < 1.5! -3.5 < -2x < -0.5! 1.75 > x > 0.25! 0.25 < x < 1.75! x!(0.25,1.75)! δ = = = 0.75 III. The ε δ Definition of imit A. Def n : et (c,d) be an open interval that containing the number a. et f be a function defined on (c,d), except possibly at a itself. Then the real number is said to be the limit of f(x) as x approaches a, denoted by lim x!a e>0, there is a δ>0 such that if 0 < x! a < ", then f (x)! < ". f ( x) =, if and only if for each B. Another way of writing the last part of this def n : f (x)! < " whenever 0 < x! a < " Note: We have replaced the imprecise descriptions (i.e. sufficiently close and arbitrarily close) in the informal definition of limits with the values epsilon and delta. C. If the value of ε is specified 1. Graphical Approach a. Use the graph of f(x)= f x if 0 < x! 8 <" then x +1! 3 <1 i.e., Use the graph of f x =3, and ε=1 to determine δ. ( ) = x +1 to find a number δ such that ( ) = x +1, -ε δ 1 = δ 2 = Now we need to find a delta neighborhood about 2 that will fit inside the open interval (3,15). x 1 a x 2 The largest delta that will work is the smaller of the 2 distances from a=8.! we choose δ <

3 2. Algebraic Approach a. Steps: 1. Solve the inequality f(x)- < ε to find an open interval (c,b) about a on which the inequality holds for all x! a ( i.e.! x " a ) 2. Find a value of δ >0 that places the open interval ( a! ",a + " ) centered at a inside the interval (a,b). The inequality f(x)- < ε will hold! x " a in this delta-interval about a. b. Example For the limit lim x! 8 that correspond to ε=1. x +1 = 3 illustrate the definition by finding the values of δ D. If the value of ε is not specified 1. An example of a proof worked out Prove that lim x! 1 ( 3x + 5) = 8 a. Preliminary analysis of the problem (guessing a value for δ) Given ε >0, find δ >0 s.t. 0 < x! a < " # f ( x)! < $ 0 < x!1 < " # ( 3x + 5)!8 < $! 3x "3 < #! 3 x "1 < #! x "1 < # 3 This suggests that we choose! = " 3 b. Proof (showing that this δ works) Given ε >0, choose! = ". If 0 < x!1 < ", then 3 ( 3x + 5 )! 8 = 3x! 3 = 3 x!1 < 3" = 3 $ # ' & % 3 ) = #. ( Thus ( 3x + 5)! 8 < " whenever 0 < x!1 < " ( 3x + 5) = 8!, by the def n of a limit, lim x! 1

4 2. Use the limit definition to prove that lim ( 4x "1) = 7. x! 2 3. Use the limit definition to prove that lim x 2 " 3x x! 5 ( ) = 10.

5 E. More Definitions 1. Def n of Right-Hand imit: et f be defined on (a,c). Then the real number is said to be the limit of f(x) as x approaches a from the right, denoted by ( ) =, if and only if for each ε>0, there is a δ>0 such that if lim f x x!a + a < x < a +!, then f (x)! < ". 2. Def n of eft-hand imit: : et f be defined on (c,a). Then the real number is said to be the limit of f(x) as x approaches a from the left, denoted by ( ) =, if and only if for each ε>0, there is a δ>0 such that if lim f x x!a " a! " < x < a, then f (x)! < ". IV. Extra Examples A. The interior of a typical 1- measuring cup is a right circular cylinder of radius 6cm. How closely must we measure the height, h, in order to measure out 1 (1000 cm 3 ) with an error of no more than 1% (i.e. 10 cm 3 )? (Use: V=πr 2 h) B. Example 2 1. Use the graph of f(x)=x 2 to find a number δ such that x 2-4 < 0.5 whenever x-2 <δ -ε δ 1 = δ 2 = Now we need to find a delta neighborhood about 2 that will fit inside the open interval 3.5, 4.5 ( ). x 1 a x 2 The largest delta that will work is the smaller of the 2 distances from x 0 =2.! we choose δ <

6 4. Now resolve the above problem using an inequality. For the limit lim x! 2 x 2 = 4 illustrate the definition by finding the values of δ that correspond to ε= Use the graph of f(x)=4x+2, =-6 and ε=1 to determine δ. x 1 a x 2 δ 1 = δ 2 = 6. Now resolve the above problem using an inequality. For the limit lim x! "2 ( 4x + 2) = "6 illustrate the definition by finding the values of δ that correspond to ε=1. -ε }δ <