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# Lesson 4: Calculating Limits (Section 21 slides)

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basic limits, limit laws, limits of algebraic and rational functions, limits of piecewise functions, limits of trigonometric quotients.

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## Lesson 4: Calculating Limits (Section 21 slides)Presentation Transcript

• Section 1.4 Calculating Limits V63.0121.021, Calculus I New York University September 16, 2010 Announcements First written homework due today (put it in the envelope) Remember to put your lecture and recitation section numbers on your paper
• Announcements First written homework due today (put it in the envelope) Remember to put your lecture and recitation section numbers on your paper V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 2 / 45
• Yoda on teaching a concepts course “You must unlearn what you have learned.” In other words, we are building up concepts and allowing ourselves only to speak in terms of what we personally have produced. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 3 / 45
• Objectives Know basic limits like lim x = a and lim c = c. x→a x→a Use the limit laws to compute elementary limits. Use algebra to simplify limits. Understand and state the Squeeze Theorem. Use the Squeeze Theorem to demonstrate a limit. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 4 / 45
• Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 5 / 45
• Heuristic Deﬁnition of a Limit Deﬁnition We write lim f (x) = L x→a and say “the limit of f (x), as x approaches a, equals L” if we can make the values of f (x) arbitrarily close to L (as close to L as we like) by taking x to be suﬃciently close to a (on either side of a) but not equal to a. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 6 / 45
• The error-tolerance game A game between two players (Dana and Emerson) to decide if a limit lim f (x) exists. x→a Step 1 Dana proposes L to be the limit. Step 2 Emerson challenges with an “error” level around L. Step 3 Dana chooses a “tolerance” level around a so that points x within that tolerance of a (not counting a itself) are taken to values y within the error level of L. If Dana cannot, Emerson wins and the limit cannot be L. Step 4 If Dana’s move is a good one, Emerson can challenge again or give up. If Emerson gives up, Dana wins and the limit is L. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 7 / 45
• The error-tolerance game L a To be legit, the part of the graph inside the blue (vertical) strip must also be inside the green (horizontal) strip. If Emerson shrinks the error, Dana can still win. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 8 / 45
• Limit FAIL: Jump y 1 x Part of graph in- −1 side blue is not inside green V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
• Limit FAIL: Jump y Part of graph in- side blue is not 1 inside green x −1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
• Limit FAIL: Jump y Part of graph in- side blue is not 1 inside green x −1 |x| So lim does not exist. x→0 x V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
• Limit FAIL: unboundedness y 1 lim+ does not exist x→0 x because the function is unbounded near 0 L? x 0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 10 / 45
• Limit EPIC FAIL π Here is a graph of the function f (x) = sin : x y 1 x −1 For every y in [−1, 1], there are inﬁnitely many points x arbitrarily close to zero where f (x) = y . So lim f (x) cannot exist. x→0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 11 / 45
• Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 12 / 45
• Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 13 / 45
• Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a Proof. The ﬁrst is tautological, the second is trivial. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 13 / 45
• ET game for f (x) = x y x
• ET game for f (x) = x y x
• ET game for f (x) = x y a x a
• ET game for f (x) = x y a x a
• ET game for f (x) = x y a x a
• ET game for f (x) = x y a x a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
• ET game for f (x) = x y a x a Setting error equal to tolerance works! V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
• ET game for f (x) = c
• ET game for f (x) = c y x
• ET game for f (x) = c y x
• ET game for f (x) = c y c x a
• ET game for f (x) = c y c x a
• ET game for f (x) = c y c x a
• ET game for f (x) = c y c x a any tolerance works! V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
• Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a Proof. The ﬁrst is tautological, the second is trivial. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 16 / 45
• Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 17 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 18 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 18 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a 2. lim [f (x) − g (x)] = L − M x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a 2. lim [f (x) − g (x)] = L − M x→a 3. lim [cf (x)] = cL x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a 2. lim [f (x) − g (x)] = L − M x→a 3. lim [cf (x)] = cL (error scales) x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
• Justiﬁcation of the scaling law errors scale: If f (x) is e away from L, then (c · f (x) − c · L) = c · (f (x) − L) = c · e That is, (c · f )(x) is c · e away from cL, So if Emerson gives us an error of 1 (for instance), Dana can use the fact that lim f (x) = L to ﬁnd a tolerance for f and g corresponding x→a to the error 1/c. Dana wins the round. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 20 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a 2. lim [f (x) − g (x)] = L − M (combination of adding and scaling) x→a 3. lim [cf (x)] = cL (error scales) x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 21 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a 2. lim [f (x) − g (x)] = L − M (combination of adding and scaling) x→a 3. lim [cf (x)] = cL (error scales) x→a 4. lim [f (x)g (x)] = L · M x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 22 / 45
• Limits and arithmetic Fact Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then x→a x→a 1. lim [f (x) + g (x)] = L + M (errors add) x→a 2. lim [f (x) − g (x)] = L − M (combination of adding and scaling) x→a 3. lim [cf (x)] = cL (error scales) x→a 4. lim [f (x)g (x)] = L · M (more complicated, but doable) x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 22 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 23 / 45
• Caution! The quotient rule for limits says that if lim g (x) = 0, then x→a f (x) limx→a f (x) lim = x→a g (x) limx→a g (x) It does NOT say that if lim g (x) = 0, then x→a f (x) lim does not exist x→a g (x) In fact, limits of quotients where numerator and denominator both tend to 0 are exactly where the magic happens. more about this later V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 24 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M n 6. lim [f (x)]n = lim f (x) x→a x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M n 6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly) x→a x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M n 6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly) x→a x→a 7. lim x n = an x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M n 6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly) x→a x→a 7. lim x n = an x→a √ √ 8. lim n x = n a x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M n 6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly) x→a x→a 7. lim x n = an (follows from 6) x→a √ √ 8. lim n x = n a x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
• Limits and arithmetic II Fact (Continued) f (x) L 5. lim = , if M = 0. x→a g (x) M n 6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly) x→a x→a 7. lim x n = an (follows from 6) x→a √ √ 8. lim n x = n a x→a n 9. lim f (x) = n lim f (x) (If n is even, we must additionally assume x→a x→a that lim f (x) > 0) x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
• Applying the limit laws Example Find lim x 2 + 2x + 4 . x→3 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
• Applying the limit laws Example Find lim x 2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: lim x 2 + 2x + 4 x→3 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
• Applying the limit laws Example Find lim x 2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
• Applying the limit laws Example Find lim x 2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 2 = lim x + 2 · lim (x) + 4 x→3 x→3 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
• Applying the limit laws Example Find lim x 2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 2 = lim x + 2 · lim (x) + 4 x→3 x→3 2 = (3) + 2 · 3 + 4 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
• Applying the limit laws Example Find lim x 2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 2 = lim x + 2 · lim (x) + 4 x→3 x→3 2 = (3) + 2 · 3 + 4 = 9 + 6 + 4 = 19. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
• Your turn Example x 2 + 2x + 4 Find lim x→3 x 3 + 11 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 27 / 45
• Your turn Example x 2 + 2x + 4 Find lim x→3 x 3 + 11 Solution 19 1 The answer is = . 38 2 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 27 / 45
• Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a) x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 28 / 45
• Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 29 / 45
• Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
• Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x→−1 x +1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
• Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x→−1 x +1 Solution x 2 + 2x + 1 Since = x + 1 whenever x = −1, and since lim x + 1 = 0, x +1 x→−1 x 2 + 2x + 1 we have lim = 0. x→−1 x +1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
• x 2 + 2x + 1 ET game for f (x) = x +1 y x −1 Even if f (−1) were something else, it would not eﬀect the limit.
• x 2 + 2x + 1 ET game for f (x) = x +1 y x −1 Even if f (−1) were something else, it would not eﬀect the limit. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 31 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 Solution We have MTP DSP lim+ f (x) = lim+ x 2 = 02 = 0 x→0 x→0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 Solution We have MTP DSP lim+ f (x) = lim+ x 2 = 02 = 0 x→0 x→0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 Solution We have MTP DSP lim+ f (x) = lim+ x 2 = 02 = 0 x→0 x→0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 Solution We have MTP DSP lim+ f (x) = lim+ x 2 = 02 = 0 x→0 x→0 Likewise: lim f (x) = lim −x = −0 = 0 x→0− x→0− V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 Solution We have MTP DSP lim+ f (x) = lim+ x 2 = 02 = 0 x→0 x→0 Likewise: lim f (x) = lim −x = −0 = 0 x→0− x→0− V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Limit of a function deﬁned piecewise at a boundary point Example Let x2 x ≥0 f (x) = −x x <0 Does lim f (x) exist? x→0 Solution We have MTP DSP lim+ f (x) = lim+ x 2 = 02 = 0 x→0 x→0 Likewise: lim f (x) = lim −x = −0 = 0 x→0− x→0− So lim f (x) = 0. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
• Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
• Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
• Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 lim = lim √ √ x→4 x − 4 x→4 ( x − 2)( x + 2) 1 1 = lim √ = x→4 x +2 4 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
• Your turn Example Let 1 − x2 x ≥1 f (x) = 2x x <1 Find lim f (x) if it exists. x→1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
• Your turn Example Let 1 − x2 x ≥1 f (x) = 2x x <1 Find lim f (x) if it exists. x→1 Solution We have DSP lim+ f (x) = lim+ 1 − x 2 = 0 x→1 x→1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
• Your turn Example Let 1 − x2 x ≥1 f (x) = 2x x <1 Find lim f (x) if it exists. 1 x→1 Solution We have DSP lim+ f (x) = lim+ 1 − x 2 = 0 x→1 x→1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
• Your turn Example Let 1 − x2 x ≥1 f (x) = 2x x <1 Find lim f (x) if it exists. 1 x→1 Solution We have DSP lim+ f (x) = lim+ 1 − x 2 = 0 x→1 x→1 DSP lim f (x) = lim (2x) = 2 x→1− x→1− V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
• Your turn Example Let 1 − x2 x ≥1 f (x) = 2x x <1 Find lim f (x) if it exists. 1 x→1 Solution We have DSP lim+ f (x) = lim+ 1 − x 2 = 0 x→1 x→1 DSP lim f (x) = lim (2x) = 2 x→1− x→1− V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
• Your turn Example Let 1 − x2 x ≥1 f (x) = 2x x <1 Find lim f (x) if it exists. 1 x→1 Solution We have DSP lim+ f (x) = lim+ 1 − x 2 = 0 x→1 x→1 DSP lim f (x) = lim (2x) = 2 x→1− x→1− The left- and right-hand limits disagree, so the limit does not exist. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
• A message from the Mathematical Grammar Police Please do not say “ lim f (x) = DNE.” Does not compute. x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
• A message from the Mathematical Grammar Police Please do not say “ lim f (x) = DNE.” Does not compute. x→a Too many verbs V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
• A message from the Mathematical Grammar Police Please do not say “ lim f (x) = DNE.” Does not compute. x→a Too many verbs Leads to FALSE limit laws like “If lim f (x) DNE and lim g (x) DNE, x→a x→a then lim (f (x) + g (x)) DNE.” x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
• Two More Important Limit Theorems Theorem If f (x) ≤ g (x) when x is near a (except possibly at a), then lim f (x) ≤ lim g (x) x→a x→a (as usual, provided these limits exist). Theorem (The Squeeze/Sandwich/Pinching Theorem) If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly at a), and lim f (x) = lim h(x) = L, x→a x→a then lim g (x) = L. x→a V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 36 / 45
• Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
• Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example π Show that lim x 2 sin = 0. x→0 x V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
• Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example π Show that lim x 2 sin = 0. x→0 x Solution We have for all x, π π −1 ≤ sin ≤ 1 =⇒ −x 2 ≤ x 2 sin ≤ x2 x x The left and right sides go to zero as x → 0. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
• Illustration of the Squeeze Theorem y h(x) = x 2 x V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
• Illustration of the Squeeze Theorem y h(x) = x 2 x f (x) = −x 2 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
• Illustration of the Squeeze Theorem y h(x) = x 2 x π 2 g (x) = x sin x f (x) = −x 2 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
• Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 39 / 45
• Two important trigonometric limits Theorem The following two limits hold: sin θ lim =1 θ→0 θ cos θ − 1 lim =0 θ→0 θ V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 40 / 45
• Proof of the Sine Limit Proof. Notice θ θ θ 1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Sine Limit Proof. Notice sin θ ≤ θ sin θ θ θ cos θ 1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Sine Limit Proof. Notice sin θ ≤ θ tan θ sin θ θ tan θ θ cos θ 1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 sin θ θ tan θ θ cos θ 1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ sin θ θ tan θ θ cos θ 1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ sin θ θ tan θ θ Take reciprocals: cos θ 1 sin θ 1≥ ≥ cos θ θ V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ sin θ θ tan θ θ Take reciprocals: cos θ 1 sin θ 1≥ ≥ cos θ θ As θ → 0, the left and right sides tend to 1. So, then, must the middle expression. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
• Proof of the Cosine Limit Proof. 1 − cos θ 1 − cos θ 1 + cos θ 1 − cos2 θ = · = θ θ 1 + cos θ θ(1 + cos θ) 2 sin θ sin θ sin θ = = · θ(1 + cos θ) θ 1 + cos θ So 1 − cos θ sin θ sin θ lim = lim · lim θ→0 θ θ→0 θ θ→0 1 + cos θ = 1 · 0 = 0. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 42 / 45
• Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 43 / 45
• Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ Answer 1. 1 2. 2 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 43 / 45
• Solutions 1. Use the basic trigonometric limit and the deﬁnition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
• Solutions 1. Use the basic trigonometric limit and the deﬁnition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 2. Change the variable: sin 2θ sin 2θ sin 2θ lim = lim 1 = 2 · lim =2·1=2 θ→0 θ 2θ→0 2θ · 2 2θ→0 2θ V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
• Solutions 1. Use the basic trigonometric limit and the deﬁnition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 2. Change the variable: sin 2θ sin 2θ sin 2θ lim = lim 1 = 2 · lim =2·1=2 θ→0 θ 2θ→0 2θ · 2 2θ→0 2θ OR use a trigonometric identity: sin 2θ 2 sin θ cos θ sin θ lim = lim = 2 · lim · lim cos θ = 2 · 1 · 1 = 2 θ→0 θ θ→0 θ θ→0 θ θ→0 V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
• Summary The limit laws allow us to compute limits reasonably. BUT we cannot make up extra laws otherwise we get into trouble. V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 45 / 45