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Definition: lim f (x) = L means: x→a (1) f is defined on an open interval containing a but not necessarily at a, and (2) for every real number ǫ > 0 there exists real number δ > 0 such that for all x in the domain of f , if 0 < |x − a| < δ, then |f (x) − L| < ǫ. In this case we say that the limit of f at a is L. Here is a graphical explanation of the definition. First of all, you have to make a clever guess at what the limit L might be. Mark this L on the y-axis. Next, we are given a positive real number ǫ – over which we have no control as we need to prove a statement for all positive ǫ. Draw a horizontal swath around y = L going ǫ units up and ǫ units down. In the picture below this swath is in light gray. L+ǫ L L−ǫ a Now the burden is on you to show that there exists a vertical swath around the vertical line x = a going δ units to the left and δ units to the right such that for any x in that swath that is not equal to a, f (x) falls into the horizontal ǫ swath. L+ǫ L L−ǫ a−δ a a+δ The thickened part of the graph now shows clearly that for all x within δ of a, f (x) falls in the given ǫ-swath. Clearly if ǫ gets smaller, δ has to get smaller too. But if ǫ gets larger, we may keep the old δ. In any case, the order of words in a definition matters. First of all, f , a, and L are given ahead of time. Then, the first-named quantity, ǫ, is chosen/given first, then δ depends on ǫ, f and a, after which xs are arbitrary depending on δ and a. Most of the time we will be giving algebraic proofs of limits, but it can help your proof if you keep in mind a geometric picture. 1 Asking for the limit of f at a is not the same thing as asking for f (a). The following three functions all have the same limit L at a; in the leftmost case, the limit equals f (a), in the middle case, the function is not defined at a, and in the rightmost case, the function is defined at a but f (a) 6= L. L L L a a a Handy facts and tricks: • • a ba b c d + dc = = ad bc . ad+bd bd . √ √ √ √ • (a − b)(a + b) = a2 − b2 . In particular, ( a − b)( a + b) = a − b. • (a + b)2 = a2 + 2ab + b2 . • (a + b)3 = a3 + 3a2 b + 3ab2 + b3 . • |ab| = |a| · |b|. • If a ≤ b then a + c ≤ b + c. If a < b then a + c < b + c. • If a ≤ b and c ≥ 0, then ac ≤ bc. • If a < b and c > 0, then ac < bc. • If a ≤ b and c < 0, then ac ≥ bc. • If a < b and c < 0, then ac > bc. • Triangle inequality: For all real numbers a and b, |a ± b| ≤ |a| + |b|. • In general, if |a| < c you cannot conclude that |a + b| < c + b. (Counterexample: a = 2, b = −5, c = 3.) • Reverse triangle inequality: For all real numbers a and b, |a ± b| ≥ |a| − |b|, and similarly, |a ± b| ≥ |b| − |a|. • Add a clever 0. • Multiply by a clever 1. • Informal goal #1 for proving epsilon-delta limits: Find “something” with the property that |f (x) − L| ≤ something · |x − a|. Informal goal #2: Find a positive real number B such that “something” from goal #1 is less than or equal to B. Then make sure that δ ≤ ǫ/B. 2 How to prove things: Statement How to prove it For all x of some type, property P holds for x. There exists x of some type such that property P holds for x. Let x be arbitrary of the specified type. Prove that property P holds for x. Find/construct/determine a concrete x of the specified type. Prove that property P holds for x. Suppose that P is true. Then prove Q. If P then Q. How does this proving scheme apply to limits? Let P be the property [of ǫ] that there exists a real number δ > 0 such that for all x in the domain of f , if 0 < |x − a| < δ, then |f (x) − L| < ǫ. This is quite a mouthful of a property P ! For proving mouthfuls, it is easier to break them into manageble pieces, and by giving them an abbreviated name, such as “P ” rather than the mouthful, it may be easier to think about it. In order to prove that every ǫ > 0 has property P , as in the chart above, the proof must start with “Let ǫ be an arbitrary positive real number.” This is somewhat verbose, and we will abbreviate it (standardly) to “Let ǫ > 0.” We did not change the meaning with this abbreviation. After this start, according to the proving chart above, we have to prove that P holds for this (arbitrary) ǫ. In other words, we have to prove that there exists a positive real δ such that et cetera. By the chart, we have to construct/find a δ first, and then prove the specifying property for δ, i.e., that for all x in the domain, et new cetera. Well, slow down, how does one construct/find a δ with the specifying property? Scratch paper comes in very handy for this, and do make use of scratch paper! But really, the best way to learn how to construct/find δ is through examples. 3 Examples: In examples below, the normal font shows a possible expected write-up, and the allcaps font is showing the thoughts that should be going on in your head at the same time and that should not be recorded in your write-up. Example 1 lim (4x − 5) = 7. x→3 Proof. The function x 7→ 4x − 5 is a polynomial function, so defined for all real numbers. [(This satisfies the first condition of limits.)] Let ǫ > 0. [(We broke up the part: For all ǫ > 0 something-or-other holds.)] Set δ = ǫ/4. [(The somethingor-other above said that there exists δ > 0 such that somethingelse holds. So, we construct an explicit δ (my experience tells me that ǫ/4 works, but you may not have such experience quite yet; read the proof in Example 2 where YOU will have to figure that out.)] Then δ is a positive real number. [(Now we have to prove that for all x, if 0 < |x − 3| < δ, then |f (x) − L| < ǫ. So: “Let x be an arbitrary real number. Assume that 0 < |x − 3| < δ.” is written in one shorter sentence next:)] Let x be any real number that satisfies 0 < |x − 3| < δ. Then |(4x − 5) − 7| = |4x − 12| = 4|x − 3| < 4δ ǫ =4 4 = ǫ. Example 2 lim (4x2 − 5x + 2) = 28. x→−2 Proof. The function x 7→ 4x2 − 5x + 2 is a polynomial function, so defined for all real [(to be determined still; your final numbers. Let ǫ > 0. Set δ = write-up will have it all filled in, but as we are writing the proof, we don’t know δ yet!)]. Then δ is a positive real number. Let x be any real number that satisfies 0 < |x + 2| < δ. Then |(4x2 − 5x + 2) − 28| = |4x2 − 5x − 26| [(x − a better be a factor of this!)] = |(x + 2)(4x − 13)| = |4x − 13| · |x + 2| [(Goal #1 accomplished: sth times |x − a|)] 4 [Goal #2: we want to bound sth above by a constant. But obviously |4x − 13| is not bounded above by a constant for all x. So we need to make a restriction on δ: let’s make sure that δ is at most 1 (or 2, or 15, it doesn’t matter what positive number you pick in this example).] [So on the δ line write: δ = min{1, = |4(x + 2 − 2) − 13| · |x + 2| [(adding a clever zero)] }.] = |4(x + 2) − 21| · |x + 2| ≤ (|4(x + 2)| + 21) · |x + 2| (by triangle inequality |a ± b| ≤ |a| + |b|) < (4 · 1 + 21) · |x + 2| (since δ ≤ 1 and |x + 2| < δ) = 25 · |x + 2| [(Goal #2 accomplished: constant times |x − a|)] [Goal #2: Now go back to δ and finish with: δ = min{1, ǫ/constant}.] < 25 · δ ≤ 25 · ǫ/25 = ǫ. Example 3 lim (4/x2 ) = 4. x→−1 Proof. The function x 7→ 4/x2 is defined for all non-zero real numbers. Let ǫ > 0. Set δ = . Then δ is a positive real number. Let x be any real number that satisfies 0 < |x + 1| < δ. Then 4 − 4x2 2 |4/x − 4| = x2 1 − x = 4 2 |1 + x|[(Goal #1 accomplished: sth times |x − a|)] x [On the δ line write: δ = min{0.4, }.] [Check that 1 in place of 0.4 would NOT work here. Any number strictly between 0 and 1 works.] 2 − (x + 1) |1 + x| = 4 x2 2 + |x + 1| ≤4 |1 + x| (by triangle inequality) x2 2.4 ≤ 4 2 |1 + x| (since δ ≤ 0.4) x 5 9.6 |1 + x| x2 9.6 |1 + x| (because x is within 0.4 of −1, hence at least 0.6 away from 0) ≤ 0.62 [On the δ line now write: δ = min{0.4, 0.62ǫ/9.6}.] 9.6 < δ 0.62 9.6 0.62 ǫ/9.6 ≤ 2 0.6 = ǫ. = Example 4 lim (x2 − 2x) = a2 − 2a. x→a Proof. Any a is a limit point of the domain of the given polynomial function. Let ǫ > 0. Set δ = min{1, ǫ/(1 + |2a − 2|)}. Let x satisfy 0 < |x − a| < δ. Then |(x2 − 2x) − (a2 − 2a)| = |(x2 − a2 ) − (2x − 2a)| (by algebra) = |(x + a)(x − a) − 2(x − a)| (by algebra) = |(x + a − 2)(x − a)| (by algebra) = |x + a − 2| |x − a| = |(x − a) + 2a − 2| |x − a| (by adding a clever 0) ≤ (|x − a| + |2a − 2|) |x − a| (by triangle inequality) ≤ (1 + |2a − 2|) |x − a| (since |x − a| < δ ≤ 1) < (1 + |2a − 2|)δ ≤ (1 + |2a − 2|)ǫ/(1 + |2a − 2|) = ǫ. 6 Exercises to go with epsilon-delta proofs and Section 1.8: 1 Fill in the blanks of the following proof that limx→2 (x2 − 3x) = −2. Explain why none of the inequalities can be changed into equalities. Let ǫ > 0. Set δ = . Let x satisfy 0 < |x − 2| < δ. Then |(x2 − 3x) − (−2)| = x2 − 3x + 2 = |x − 1| |x − 2| (because < (3 + 1) |x − 2| (because ≤ (|x − 2| + 1) |x − 2| < 4δ ǫ ≤4 4 = ǫ. ) ) (because ) (because ) (because ) 2 Determine the following limits and prove them with epsilon-delta proofs. i) lim (x3 − 4). x→1 1 . x→2 x lim xx−4 2 +2 . x→3 √ ii) lim iii) iv) lim x→4 x + 5. x2 −9 . x→3 x−3 lim x−3 . 2 x→3 x −9 v) lim vi) 3 Let b be a real number and let f, g be functions defined for all real numbers a follows: f (x) = x3 − 4x2 , b, if x 6= 5; if x = 5, g(x) = x3 − 4x2 , b, if x = 5; if x = 6 5. Prove that the limit of f (x) as x approaches 5 is independent of b, but that the limit of g(x) as x approaches 5 depends on b. 4 Prove that limx→a (mx + l) = ma + l, where m and l are constants. When a limit is not L Suppose that f is defined on an open interval containing a but not necessarily at a. Then lim f (x) 6= L means: x→a NOT For all real numbers ǫ > 0 there exists a real number δ > 0 such that for all x in the domain of f , if 0 < |x − a| < δ then |f (x) − L| < ǫ. = There exists a real number ǫ > 0 such that NOT there exists a real number δ > 0 such that for all x in the domain of f , if 0 < |x − a| < δ then |f (x) − L| < ǫ. = There exists a real number ǫ > 0 such that for all real numbers δ > 0, NOT for all x in the domain of f , if 0 < |x − a| < δ then |f (x) − L| < ǫ. = There exists a real number ǫ > 0 such that for all real numbers δ > 0, there exists x in the domain of f such that NOT if 0 < |x − a| < δ then |f (x) − L| < ǫ. = There exists a real number ǫ > 0 such that for all real numbers δ > 0, there exists x in the domain of f such that 0 < |x − a| < δ and NOT |f (x) − L| < ǫ. = There exists a real number ǫ > 0 such that for all real numbers δ > 0, there exists x in the domain of f such that 0 < |x − a| < δ and |f (x) − L| ≥ ǫ. x Example 1 The limit of |x| as x approaches 0 does not exist. In other words, for all real x numbers L, limx→0 |x| 6= L. x Proof. The domain of the function that takes x to |x| is the set of all non-zero real numbers, so that 0 is a limit point of the domain. Set ǫ = 1 (Hint: half of the gap). Let δ > 0 be an arbitrary positive number. Let x = −δ/2 if L ≥ 0, and let x = δ/2 otherwise. In both cases, 0 < |x| = |x − 0| = δ/2 < δ. If L ≥ 0, then −δ/2 x |x| − L = | − δ/2| − L = | − 1 − L| = 1 + L ≥ 1 = ǫ, and if L < 0, then x δ/2 = = |1 − L| = 1 − L > 1 = ǫ. − L − L |x| |δ/2| This proves the claim of the example. Example 2 For f be given by the graph below, lim f (x) does not exist because of the x→2 jump in the function at 2. 2 1 1 2 3 Here is an epsilon-delta proof. Say that the limit exists. Call it L. Set ǫ = 41 . Let δ be an arbitrary positive number. If L ≥ 32 , set x = 2 + min 14 , 2δ , and if L < 32 , set x = 2 − min{ 14 , 2δ }. In either case, 0 < |x − 2| = min 1 δ , 4 2 ≤ δ < δ. 2 If L ≥ 23 , by our choice x = 2 + min{ 41 , δ2 }, so that f (x) = 1 + min{ 41 , δ2 } ≤ 1 + 41 , whence |f (x) − L| ≥ 41 = ǫ. Similarly, if L < 23 , by our choice x = 2 − min{ 41 , δ2 }, so that f (x) = 2 − min{ 41 , 2δ } ≥ 2 − 14 = 1 + 34 , whence |f (x) − L| ≥ 41 = ǫ. Thus no L works, so the limit of f (x) as x approaches 2 does not exist. Exercise 3 Below you will find an attempt at a “proof” that limx→3 x2 = 9. Explain how the two starred steps below contain three mistakes. Let ǫ > 0. Set δ =∗ ǫ/|x + 3|. Then |x2 − 9| = |x − 3| · |x + 3| <∗ δ|x + 3| = ǫ. Definition: The right-sided limit of f at a is L: lim f (x) = L means: x→a+ (1) f is defined on an open interval of the form (a, b) for some b > a, and (2) for every real number ǫ > 0 there exists real number δ > 0 such that for all x in the domain of f , if 0 < x − a < δ, then |f (x) − L| < ǫ. Definition: The left-sided limit of f at a is L: lim f (x) = L means: x→a− (3) f is defined on an open interval of the form (a, b) for some b > a, and (4) for every real number ǫ > 0 there exists real number δ > 0 such that for all x in the domain of f , if 0 < a − x < δ, then |f (x) − L| < ǫ. Example 4 lim+ x→3 √ 2x − 6 = 0. Proof. The domain here is all x ≥ 3. So 3 is a limit point of the domain. Let ǫ > 0. Set δ = ǫ2 /2. Let x satisfy 0 < x − 3 < δ. Then √ 2x − 6 = < = √ √ √ 2· 2· √ √ x−3 δ p 2 · ǫ2 /2 = ǫ. x2 + 4, Example 5 Let f be given by f (x) = x − 2, lim f (x) = −3. if x > 1; if x ≤ 1. Then lim+ f (x) = 5, x→1− Proof. Let ǫ > 0. Set δ = min{1, ǫ/3}. Let x satisfy 0 < x − 1 < δ. Then f (x) − 5 = x2 + 4 − 5 (since x > 1) = x2 − 1 = (x + 1)(x − 1) < (x + 1)δ (since x > 1, so x + 1 is positive) = (x − 1 + 2)δ (by rewriting) < (1 + 2)δ (since x − 1 < δ ≤ 1) ≤ 3ǫ/3 = ǫ. This proves that lim f (x) = 5. x→1+ x→1 Set δ = ǫ. Let x satisfy 0 < 1 − x < δ. Then f (x) − (−3) = x − 2 + 3 (since x < 1) =x−1 <δ = ǫ. This proves that lim f (x) = −3. x→1− Exercises to go with one-sided limits: 1 Let f be given by f (x) = x |x| . Prove that limx→0+ f (x) = 1 and that limx→0− f (x) = −1. 2 Find a function f defined for all real numbers such that limx→0− f (x) = 2 and such that limx→0+ f (x) = −5. 3 Find a function f defined for all real numbers such that limx→0− f (x) = 2, limx→0+ f (x) = −5, limx→1− f (x) = 3, and limx→1+ f (x) = 0. (Try to define such a function with fewest possible words or symbols, but do use full grammatical sentences.)