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MAC 2313LIMITS AND CONTINUITYI. LIMITS. In this section, we will keep in mind two limit definitions:(i) ² − δ definition from the book, and(ii) “sequence” definition:Suppose that f (x, y) is defined in a neighborhood of (x0, y0), exceptpossibly at (x0, y0). We say thatlim(x,y)→(x0,y0)f(x, y) = Lif for every sequence (xn, yn) of points (in the domain of f) such that(xn, yn) → (x0, y0), we have f(xn, yn) → L, as n → ∞.It is sometimes convenient to use (ii) to show that the limit of f at(x0, y0) does not exist. You can do this, for instance, by finding twosequences (xn, yn) and (an, bn) of points (in the domain of f ) such that(xn, yn) → (x0, y0) and (an, bn) → (x0, y0), but f(xn, yn) and f(an, bn)converge to different values. Here is an example.Example 1. Does the limit(1) lim(x,y)→(0,0)2xyx2+ y2exist?Solution: Consider sequences (1n,1n) and (1n,2n). Both sequences clearlyconverge to (0, 0). However,fµ1n,1n¶=2n22n2→ 1 and fµ1n,2n¶=4n25n2→45.Hence, the limit (1) does not exist.Another way to see this is to consider the limits of f as (x, y) →(0, 0) along two straight lines (or curves) and show that the results aredifferent. For instance, finding the limits along y = x and then alongy = 2x, you get the answers 1 and45respectively. Different answersmean that the limit (1) does not exist.Exercise 1. Show that the following limit does not exist:lim(x,y)→(0,0)x2y2x2y2+ (x − y)212Exercise 2. Find the following limit:lim(x,y)→(+∞,+∞)x + yx2−xy+y2.Outline: We will use the “squeeze theorem,” which works the sameway as in Calculus 1.Our goal is to make the following estimate (which works for all x 6= 0and y 6= 0):(2) 0 ≤¯¯¯¯x + yx2− xy + y2¯¯¯¯≤1|y|+1|x|.Clearly, the rightmost side of (2) goes to 0 as (x, y) → (+∞, +∞),and the leftmost side of (2) is 0. Hence, by “squeeze theorem” themiddle term must go to 0.Let’s make the estimate (2). The first inequality in (2) is easy:absolute value must be greater than or equal to zero. To prove thesecond inequality, we begin by showing that for all numbers x and ywe havex2− xy + y2≥ xy.Then, explain why it follows that|x2− xy + y2| ≥ |xy|.Thus, for all x and y (with x 6= 0 and y 6= 0) we get0 ≤¯¯¯¯x + yx2− xy + y2¯¯¯¯≤¯¯¯¯x + yxy¯¯¯¯=¯¯¯¯1x+1y¯¯¯¯≤1|x|+1|y|.Explain why these inequalities are true. (For instance, notice thatthe last inequality follows from the celebrated triangle inequality fornumbers: |u + v| ≤ |u| + |v|).Exercise 3. Show thatlim(x,y)→(+∞,+∞)x2+ y2x4+ y4= 0.Hint: Using the kind of reasoning as in Exercise 2, show that for allx 6= 0 and y 6= 0 we have:0 <x2+ y2x4+ y4≤1x2+1y2.Then use the “squeeze theorem.”Exercise 4. Show thatlim(x,y)→(0,0)(x2+ y2)x2y2= 1.3Outline: First, since we are considering the limit as (x, y) → (0, 0),it is enough to consider just those points that are sufficiently close to(0, 0), say points 0 < x2+ y2< 1.Our goal will be to make the following estimate for 0 < x2+ y2< 1:(3) (x2+ y2)(x2+y2)24≤ (x2+ y2)x2y2≤ 1.Now, to explain (3), we first show that for any x and y we have:x2y2≤14(x2+ y2)2,and, then, to finish the explanation, we use the fact that we are con-sidering only 0 < x2+ y2< 1.Having done this, we try to use the squeeze theorem. For this, weneed to show thatlim(x,y)→(0,0)(x2+ y2)(x2+y2)24= 1.This does not look like an easy task. However, we can make it easierby passing to polar coordinates. Indeed, this will be very nice as ourfunction depends only on the radius r =px2+ y2:lim(x,y)→(0,0)(x2+ y2)(x2+y2)24= limr→0+(r2)r44.Now use the “00” case of L’Hopital’s rule, and you will get 1.II. CONTINUITYExample 2. Briefly explain why the functionf(x, y) =½2xyx2+y2, (x, y) 6= (0, 0)0, (x, y) = (0, 0).is continuous at all points (x, y) 6= (0, 0). Why is f discontinuous at(0, 0)?Solution: As a quotient of two polynomial functions, f is continuousat all points ( x, y) where the denominator is not equal to 0, i.e. for all(x, y) 6= (0, 0).Recall from Example 1 that the limitlim(x,y)→(0,0)2xyx2+ y2does not exist. Hence, f is not continuous at (0, 0).4Exercise 5. Briefly explain why the functionf(x, y) =½x2yx4+y2, (x, y) 6= (0, 0)0, (x, y) = (0, 0).is continuous at all points (x, y) 6= (0, 0). Why is f discontinuous at(0, 0)?Exercise 6. (i) Show that the functionf(x, y) =½x4x2+y2, (x, y) 6= (0, 0)2, (x, y) = (0, 0).is continuous at all points (x, y ) 6= (0, 0).(ii) Redefine the function f at (0, 0) so that the new function is con-tinuous at (0, 0).Outline: As in previous exercises, explain why f must be continuousfor all (x, y) 6= (0, 0).Then, consider the point (0, 0). Clearly, f is defined at (0, 0) withthe value f(0, 0) = 2. Unlike in Example 2 above, the limitlim(x,y)→(0,0)x4x2+ y2exists and is equal to 0. Explain why this is true. For instance, youmay pass to polar coordinates or use “squeeze theorem.”This, however, does not mean that f is continuous at (0, 0). In fact,from the definition of f we know that f(0, 0) = 2, and we have justfound out the value of the limit is 0. Hence, f is not continuous at(0, 0).So, how can we redefine f at (0, 0) so that the new function is con-tinuous at (0, 0)?The kind of discontinuity described in this example is known as a“removable discontinuity.”Exercise 7. (i) Show that the following function is defined for all(x, y) 6= (0, 1):f(x, y) =x(y − 1)(x2− y2+ 2y − 1)x2+ y2− 2y + 1.(ii) Show thatlim(x,y)→(0,1)x(y − 1)(x2− y2+ 2y − 1)x2+ y2− 2y + 1= 05(iii) Extend the definition of f so that the new function is continuousat (0, 1).Hint: Show that f can be written asf(x, y) =x(y − 1)(x2− (y − 1)2)x2+ (y − 1)2.This will answer (i).To answer (ii), make a polar-type change of coordinates x = r cos θand y − 1 = r sin θ. Explain why (x, y) → (0, 1) means that r → 0+.Compute the limit in new coordinates. Now you can answer (iii).Exercise 8. (i) Show that the functionf(x, y) =x + yx3+ y3is defined and continuous at all points (x, y ) such that x + y 6= 0.(ii) Consider the points (x0, y0) such that (x0, y0) 6= (0, 0) and x0+y0=0. Show that at these points,lim(x,y)→(x0,y0)x + yx3+ y3=1x20− x0y0+ y20.Explain why the quantity on the right hand side is finite.(iii) Show thatlim(x,y)→(0,0)x + yx3+ y3= +∞.(iv) Explain why discontinuities in (ii) are removable, and why the onein (iii) is not removable.Example 3. Using ²-δ definition show that f(x, y) = x + y is contin-uous at all points (x0, y0)


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