October 27, 208 MAT86 Week 3 Justin Ko Limits. Intuitive Definitions of Limits We use the following notation to describe the iting behavior of functions.. (Limit of a Function A it is written as f( = L a the it of f( as approaches a is L. This means f gets arbitrarily close to L whenever is sufficiently close but not equal to a. 2. (One-sided it A right one-sided it is written as f( = L a + the it of f( as approaches a from the right is L. This means f gets arbitrarily close to L whenever is sufficiently close but strictly greater than a. Similarly, a left one-sided it is written as f( = L a the it of f( as approaches a from the left is L. This means f gets arbitrarily close to L whenever is sufficiently close but strictly less than than a. 3. (Limit at Infinity A it at infinity is written as f( = L the it of f( at infinity is L. This means f gets arbitrarily close to L whenever is sufficiently large. Similarly, a it at negative infinity is written as f( = L the it of f( at negative infinity is L. This means f gets arbitrarily close to L whenever is sufficiently small. 4. (Infinite Limit If L = + in the definitions above, then it means that f gets arbitrarily large. Likewise, if L = then we mean that f gets arbitrarily small..2 Special Limits We have the following special its. 2. 3. 4. sin 0 =, cos = 0, 0 ( ( + a = e a or + a n = e 0 n ± n a, e =. 0 Page of 5
October 27, 208 MAT86 Week 3 Justin Ko.3 Limit Laws Suppose that f( and g( eists. a a We can use the following properties to compute the its of complicated functions.. Sums of Functions: 2. Products of Functions: [f( + g(] = f( + g(. a a a [f( g(] = f( g(. a a a 3. Direct Substitution: If f is continuous at a, then f( = f(a. a 4. Composition Theorem: If f( is continuous at L and a g( = L, then f(g( = f( g( = f(l. a a 5. Squeeze Theorem: If g( f( h( when is near a (ecept possibly at a and then g( = h( = L a a f( = L. a Remark: The functions introduced in Week, i.e. polynomials, eponentials, trigonometric functions, hyperbolic functions, and their inverses, etc are continuous at every number in their domains..4 Infinite Limits and Graphs Limits at infinity or approaching infinity tell us a lot about the shape of our graph. Consider the curve y = f(, there are three types of typical behavior described by these its. Vertical Asymptote: A vertical line = a is called a vertical asymptote of the curve y = f( if at least one of the following occurs, f( = ±, or a A curve may have multiple vertical asymptotes. f( = ±, or f( = ±. a + a 2. Horizontal Asymptote: A horizontal line y = L is called a horizontal asymptote of the curve y = f( if [f( L] = 0, or [f( L] = 0. Note that a curve may have 0, or 2 horizontal asymptotes. 3. Slant Asymptote: An oblique line y = m + b (we require m 0 is called a slant asymptote of the curve y = f( if [f( (m + b] = 0 or Note that a curve may have 0, or 2 slant asymptotes. [f( (m + b] = 0. Page 2 of 5
October 27, 208 MAT86 Week 3 Justin Ko 0 5 8 7 6 5 4 3 2 0 2 3 4 5 6 7 8 5 0 Eample : The graph of + > y = f( = 3 2 + < 4( is displayed above. This graph has a vertical asymptote at =, a horizontal asymptote y = as approaches, and a slant asymptote y = 3 2 as approaches..5 Eample Problems.5. Showing a Limit Eists Strategy: In this course, we do not have to use the definition of a it to show a it eists. Instead, it suffices to use an algebraic approach and it laws to derive a it. The following procedure generally works for combinations of functions we encountered in Week,. We first check f(a to figure out the form of the it. If this gives us a number, then f(a is our it and there is nothing more to do (provided that we are dealing with the composition of continuous functions. 2. If we have an indeterminate form, 0 0,, 0,, 00,, 0, etc then we simplify our function by factoring, collecting like terms, rationalizing, etc. The goal is to remove the bad point that makes our behavior indeterminate. 3. After simplifying our function, we evaluate our simplified function at a. If it gives us a number, then that is our it. Useful Formulas: When factoring, the following formulas may be useful. Difference of Powers: For a, b R and n 2, (a n b n = (a b(a n + a n 2 b + + ab n 2 + b n. There are n terms in the second part of the factored equation. For eample, if n = 4, then ( 4 y 4 = ( y( 3 + 2 y + y 2 + y 3. Page 3 of 5
October 27, 208 MAT86 Week 3 Justin Ko 2. Quadratic Formulas: For a 0, the equation a 2 + b + c = 0 has solution(s = b ± b 2 4ac. 2a Problem. ( Determine the following it 4 2 + 4. + 4 Solution. We first try to substitute = 4 into our function 2 + 4 + 4 This is not an indeterminate form, so we can conclude = 42 + 4 4 = 2. =4 4 + 4 2 + 4 = 2. 4 + 4 Problem 2. ( Determine the following it 2 +. Solution 2. We first try to substitute = into our function 2 + = ( 2 = 0 = + 0. This is an indeterminate form, so we attempt to simplify our function by factoring. Using the difference of squares formula to factor the numerator, we notice 2 + = ( ( + = ( = 2. + Problem 3. ( Determine the following it 5 3. Solution 3. We first try to substitute = into our function 5 3 = 5 = 3 = 0 0. This is an indeterminate form, so we attempt to simplify our function by factoring. Using the difference of powers formula to factor the numerator and denominator, we notice 5 3 = ( ( 4 + 3 + 2 + + 4 + 3 + 2 + + ( ( 2 = + + 2 = 5 + + 3. Page 4 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem 4. ( Determine the following it 0 2 + 2. Solution 4. We first try to substitute = into our function 2 + =0 2 = = 0 0 0. This is an indeterminate form, so we attempt to simplify our function by rationalizing the numerator. We notice 2 + 2 + 2 + + 0 2 = 0 2 2 + + ( 2 + 2 = 0 2 ( 2 + + = 0 2 + + = 2..5.2 Showing a One-Sided Limit Eists Strategy: The strategy is identical as the last section. We might have to be careful with piecewise functions in these eamples. Problem. ( Determine the following it 7 0. Solution. We first try to substitute = 0 into our function 7 = 7 0 = 0 =0 0 0. This is an indeterminate form, so we attempt to simplify our function by canceling the numerator. Since = for < 0, we have 7 0 = 7 0 = 7 = 7. 0 Problem 2. ( Determine the following it ( + 2 2. Solution 2. We first try to substitute = + into our function ( 2 = 2 =. + Page 5 of 5
October 27, 208 MAT86 Week 3 Justin Ko This is an indeterminate form, so we attempt to simplify our function by collecting like terms. We notice ( + 2 ( ( + 2 = + ( ( + 2 ( ( + = + ( ( + = + + = 2..5.3 Showing a Limit Does not Eist Strategy: Unless our function is eceptionally ugly, the usual way to show a it does not eist is to show that its left and right its are different. Problem. ( Show that the following it does not eist, 3 3 3. Solution. It suffices to show that our left and right its approach different values. Since = for > 0, we have 3 3 + 3 = 3 3 + 3 = =, 3 + and since = for < 0, we have 3 3 3 = 3 3 ( 3 = =. 3 The left and right its do not agree, so our it does not eist. Problem 2. ( Show that the following it does not eist, sin(. 0 + Solution 2. Let t = /. We have 0 + f( = t f( t. In particular, we have sin( = sin(t 0 + t which does not have a it, because sin(t oscillates between and infinitely often and does not converge to a particular value..5.4 Limits of the composition of functions If f is discontinuous at b and a g( = b, then a f(g( may not behave nicely. These eamples are to illustrate the potential dangers when taking its with a discontinuous function on the outside. In these cases, evaluating the function at a may not give the right answer even if we do not get an indeterminate form. Page 6 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem. ( Consider the discontinuous function { = 0 f( = 0 0 and let g( = 2. Compute 0 f(g(. Solution. We first compute the composition of the functions, { f(g( = f( 2 = 0 = 0 0. since 2 = 0 if and only if = 0. Therefore, we have since f(g( is equal to 0 for all 0. f(g( = 0, 0 Note: We cannot take the it inside function f in this problem. In this case, if we incorrectly used the property 0 f(g( = f( 0 g( then we will incorrectly conclude f(g( = 0 0 f(2 = f( 2 = f(0 =. 0 The reason why we can t take the it inside is that f( is not continuous at 0, so the property does not hold. This is a reason why eistence of the its is not enough for the composition of functions. Problem 2. ( Consider the discontinuous function { > 0 f( = 0 0 and let g( = 2. Compute 0 f(g(. Solution 2. We first compute the composition of the functions, { f(g( = f( 2 0 = 0 = 0. since 2 > 0 whenever 0 and 2 = 0 when = 0. Therefore, we have since f(g( is equal to for all 0. f(g( =, 0 Note: In this eample, even though the the it 0 f( does not eist (check that the left and right its give different values, the it of f(g( still eists. If we tried to take the it inside, 0 f(g( = f( 0 g( then we would have gotten the incorrect answer, for the same reasoning as the previous problem. Page 7 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem 3. ( Consider the discontinuous function { = 0 f( = 0 0. Compute 0 f(f(. Solution 3. We first compute the composition of the functions, { 0 = 0 f(f( = 0. Therefore, we have f(f( =. 0 Note: In this case, if we incorrectly used the property 0 f(g( = f( 0 g( then we get f(f( = f( f( = f(0 =, 0 0 which is the correct answer. This just happened by chance, and the steps to arrive at this conclusion is completely wrong without more justification. This is also the usual counter eample that disproves the claim that if g b f(g = L and a g( = b, then a f(g( = g b f(g = L..5.5 Limits at Infinity (Horizontal Asymptotes Strategy: The it as is usually straightforward. We proceed as follows:. We first check f( to figure out the form of the it. If this gives us a number, then f( is our it and there is nothing more to do. 2. If our function is in an in-determinant form, then we simplify our equation until it is of the form or 0 0. 3. Divide the numerator and denominator by the highest degree term, then take and recompute the it. If we take the and there is a square root in our problem, then we have to be careful. It is usually best to make a change of variables t =, and use the fact f( = t f( t. Problem. ( Determine the following it +2 e 2. Solution. We first try to substitute into our function e 2 +2 = e 2 +2 = e = 0 = is not an indeterminate form, so +2 e 2 = 0. Page 8 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem 2. ( Determine the following it + 2 + 2 + 7. Solution 2. We first try to substitute into our function + = 2 + 2 + 7 =. This is an indeterminate form of the form, so we can divide the numerator and denominator by the highest degree polynomial, + 2 + 2 + 7 = + 2 + 2 + 7 = + =. + 2 + 7 2 Problem 3. ( Determine the following it + 2 + 2 + 7. Solution 3. We first try to substitute into our function + = 2 + 2 + 7 =. This is an indeterminate form of the form, so we can divide the numerator and denominator by the highest degree polynomial. Since there is a square root in our question, we have to be careful because we cannot put a negative number inside of the square root to simplify it. Instead, we use the change of variables t = and notice + 2 + 2 + 7 = t t + ( t2 2t + 7 = t t t t + t2 2t + 7 = + t =. t 2 t + 7 t 2 Problem 4. ( Determine the following it ( 9 2 + 3. Solution 4. We first try to substitute into our function ( 9 2 + 3 =. = This is an indeterminate form of the form. This form is really hard to work with, so we first rewrite our function by rationalizing the root ( 9 2 + 3 = ( 92 9 2 + + 3 + 3 92 + + 3 = 92 + + 3. It is easy to check that this it is of the form, so we can divide the numerator and denominator by the highest power and conclude 92 + + 3 = 92 + + 3 = 9 + + 3 = 9 + 3 = 6. Page 9 of 5
October 27, 208 MAT86 Week 3 Justin Ko.5.6 Infinite Limits (Vertical Asymptotes Strategy: To find the behavior of our function at a vertical asymptote, we first substitute a value of at the asymptote, and remember the asymptotic behavior of the graphs of the standard function 0 + =, 0 =, e =, ln(0 + =, ln( =, tan( π =, etc 2 If our answer is infinite, then we don t need to do any more work. However, if we have an indeterminate form, then we can proceed like usual to compute the it. If we want to find the vertical asymptotes of a combination of basic functions introduced, we should start by checking for asymptotes where f( is undefined. Problem. ( Determine the following it ( 2. ln( Solution. We first substitute = into our function 2 ln( = 2 = ln( = 2 0 = 2 + =. This is not an indeterminate form, so we can conclude, ( 2 =. ln( Problem 2. ( Find the vertical asymptotes of the function f( = + 2 4. Solution 2. Since f( is undefined when = ±2, and continuous everywhere else, it suffices to check for vertical asymptotes at = ±2. Notice that + 2 + 2 = 3 + = and 0 + 2 2 = 3 0 = so there is vertical asymptote at = 2. Similarly, + 2 + 2 = + = and 0 2 2 = 0 + = so there is vertical asymptote at = 2..5.7 Special case of a it at infinity (Slant Asymptotes Recall that a line y = m + b and m 0 is called a slant asymptote if [f( (m + b] = 0 or [f( (m + b] = 0. We call the line m+b a slant asymptote at (or a slant asymptote at if the second case occurs. Page 0 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem. ( Find the slant asymptotes of the function f( = 3 2 +. Solution. Using long division, we notice that 3 2 + = 2 +. In this form, we see that f( is almost the equation of a line + b, with the intercept term b = going to 0 as gets large. To check that is a slant asymptote at, we have [ 3 ] 2 + = 2 + = 0 and similarly, to check that is a slant asymptote at, we have [ 3 ] 2 + = 2 + = 0. 2 + Problem 2. ( Find the slant asymptotes of the function f( = π + tan (. Solution 2. We see that f( is almost the equation of a line π + b, with the intercept term b = tan ( going to π/2 as gets large and π/2 as gets small. To check that π + π/2 is a slant asymptote at, we have [π + tan ( (π + π 2 ] = tan ( π 2 = 0 and similarly, to check that π π/2 is a slant asymptote at, we have [π + tan ( (π π ] 2 = tan ( + π 2 = 0. The graph of the f( is in red and the slant asymptotes are plotted in the dotted blue lines, Page of 5
October 27, 208 MAT86 Week 3 Justin Ko.5.8 Squeeze Theorem Problems Strategy: We want to find upper and lower bounds of f( that have the same its as a. Problem. ( Determine the following it ( sin. 0 + Solution. We first substitute = 0 + into our function ( sin = 0 sin(. =0+ which is an indeterminate form. There is no algebraic way of simplifying the sin(, so we use the Squeeze theorem to find an appropriate it. Notice that the range of sin( [, ], so we have ( sin for all > 0. Therefore, multiplying both sides by > 0, we have ( sin Computing the its of the lower and upper bounds we see 0 +( = 0 and = 0. 0 + These its are the same, so the squeeze theorem implies ( sin = 0. 0 + Problem 2. ( Determine the following it e sin(π/. 0 + Solution 2. We first substitute = 0 + into our function e sin(π/ = 0 e sin(. =0 + which is an indeterminate form. There is no algebraic way of simplifying the e sin(π/, so we use the Squeeze theorem to find an appropriate it. Notice that the range of sin( [, ], so we have ( π sin for all > 0. Therefore, eponentiating everything we have e e sin(π/ e and then multiplying by, we have e e sin(π/ e. Computing the its of the lower and upper bounds we see e = 0 and 0 + These its are the same, so the squeeze theorem implies e = 0. 0 + e sin(π/ = 0. 0 + Page 2 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem 3. ( Prove the basic trigonometric it, sin( =. 0 Solution 3. Let 0 < < π 2 and consider the picture of the unit circle: B D sin( tan( O cos( C A From the picture, we see that the area of the inner triangle OCB is less than the area of the sector OAB, which is less than the area of the outer triangle OAD. Computing the areas eplicitly using the formula for the triangle and the sector of a circle, we have 2 sin( cos( 2 tan( sin( cos( tan(. 2 Since sin( > 0 on this interval, we can divide by sin( and conclude cos( sin( cos(. Taking reciprocals (the direction of the inequalities reverse, we have cos( sin( cos(. Computing the its as 0 + of the upper and lower bounds, we see =, and 0 + cos( These its are the same, so the squeeze theorem implies sin( =. 0 + cos( =. 0 + To compute the other one-sided it, notice that sin( is even, so sin( 0 = by symmetry. Page 3 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem 4. ( Determine the following it Hint: Use the fact sin(θ θ for all θ R. sin( 2 + 0 + 4 + 4. 2 Solution 4. We first substitute = 0 + into our function sin( 2 + 4 + 4 2 =0 + = 0 0. which is an indeterminate form. For θ 0, we have the bounds θ sin(θ θ (we can prove this bound later in the course using the mean value theorem. Taking θ = 2, for > 0 we have 2 sin( 2 2. Therefore, we can add and divide by 4 + 4 2 to conclude 2 + 4 + 4 sin(2 + 2 4 + 4 2 + 2 4 + 4. 2 Computing the its of the lower and upper bounds, we see and 0 + 0 + 2 + 4 + 4 = + 2 0 + 2 + 4 = = 4 2 2 + 4 + 4 = + 2 0 + 2 + 4 = = 4 2. These its are the same, so the squeeze theorem implies sin( 2 + 0 + 4 + 4 = 2 2. Note: The identity 0 sin(θ θ for θ > 0 can also be used to derive the same it. However, one can check that the usual bounds sin( will not work in this problem, so we have to use a sharper bound for sin(..5.9 Using Special Limits Problem. ( Determine the following it sin( 2 + 0 + 4 + 4. 2 Solution. Instead of using the squeeze theorem like in the last section, we can use the identity =. In this case, if we multiply the numerator and denominator by, we have 0 sin( 0 + sin(2 + 4 + 4 2 = 0 + sin( 2 + 2 + 4 = 0 + sin( 2 + 2 = = 2 + 4 4 2. Page 4 of 5
October 27, 208 MAT86 Week 3 Justin Ko Problem 2. ( Determine the following it 0 sin(3 tan( sin 3. (9 sin( Solution 2. We can use the identity 0 = to solve this problem. Since our functions are continuous, we can use the it composition theorem to conclude for all a 0, sin(a sin(a sin(t = a = a = a and 0 0 a t 0 t 0 sin(a = 0 where we used the fact that 0 is the same as t = a 0. Therefore, sin(3 tan( sin(3 sin( 0 sin 3 = (9 0 sin 3 (9 cos( ( = 0 sin(9 sin(3 sin(9 = 0 ( sin(9 sin(3 sin( sin(9 cos( sin(9 sin( = 9 3 9 9 sin(a = a 0 = 243. ( sin(a sin(9 = a cos( Problem 3. ( Determine the following it 0 cos(3. sin(2 sin( Solution 3. We can use the identity 0 We have, cos(3 ( = cos(3 0 sin(2 0 ( = 3 0 2 cos(3 3 = 3 2 0 2 = 0. = and 0 cos( sin(2 2 sin(2 sin(a cos(a =, = 0 0 a 0 a = 0 to solve this problem. Page 5 of 5