Questions and Answers #4 Limits and Continuity

Questions and Answers Sheet 4 Limits and Continuity Question #1 Discuss the continuity of the functions f and g. Explain any differences  x 2 + 2 xy 2 + y 2 , ( x, y )  0  f ( x, y ) =  x2 + y 2 0, ( x, y ) = 0   x 2 + 2 xy 2 + y 2 , ( x, y )  0  g ( x, y ) =  x2 + y 2 1, ( x, y ) = 0  Answer: We will first consider the given functions, The objective us to discuss the continuity of the functions. First, we will consider But f(0,0)=0 Now, as lim lim 𝑥 2 +2𝑥𝑦 2 +𝑦 2 (𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2 = 0+2(0)+𝑦 2 lim (0,𝑦)→(0,0) 0+𝑦 2 𝑓(𝑥, 𝑦) ≠ 𝑓(0,0) (𝑥,𝑦)→(0,0) Hence, we can say that the function f(x,y) is continuous everywhere except at (0,0). Next, we will consider  x 2 + 2 xy 2 + y 2 , ( x, y)  0  g ( x, y ) =  x2 + y 2 1, ( x, y) = 0  Now, we will find lim (𝑥,𝑦)→(0,0) lim (𝑥,𝑦)→(0,0) 𝑥 2 +2𝑥𝑦 2 +𝑦 2 𝑥 2 +𝑦 2 =1 And g(0,0)=1 Now, as lim (𝑥,𝑦)→(0,0) = lim 𝑔(𝑥, 𝑦) along the kine x=0, we get, (0,𝑦)→(0,0) 0+2(0)+𝑦 2 0+𝑦 2 𝑔(𝑥, 𝑦) = 𝑔(0,0) Hence, we can say that the function g(x,y) is continuous everywhere. Question #2 Continuity and limts: Show a function is continuous at a 𝑓(𝑥) = (𝑥 + 2𝑥 3 )4 , 𝑎 = −1 Answer: Show a function is continuous at a 𝑓(𝑥) = (𝑥 + 2𝑥 3 )4 , 𝑎 = −1 We have 𝑓(𝑥) = (𝑥 + 2𝑥 3 )4 , 𝑎 = −1 𝑓(𝑥) = (𝑥 + 2𝑥 3 )4 𝑎𝑡 𝑎 = −1 to be continous if, L.H.L=R.H.L=Function at the given point Now, 3 4 lim 𝑓(𝑥) = lim (𝑥 + 2𝑥 3 )4 = ( lim 𝑥 + 2 ( lim 𝑥) ) 𝑥→−1 𝑥→−1 𝑥→−1 𝑥→−1 ⇒ lim 𝑓(𝑥) = (−1 + 2(−1)3 )4 = (−1 − 2)4 = (−3)4 𝑥→−1 ⇒ lim 𝑓(𝑥) = 81 𝑥→−1 ∴ 𝐿. 𝐻. 𝐿 = lim 𝑓(𝑥) = 81 𝑥→−1 Then, lim 𝑓(𝑥) = (1 + 2(1)3 )5 = (1 + 2)4 = (3)4 𝑥→1 ∴ 𝑅. 𝐻. 𝐿 = lim 𝑓(𝑥) = 81 𝑥→1 Now, for f(-1) 4 ⇒ 𝑓(−1) = ((−1) + 2(−1)3 ) = (−1 − 2)4 = (−3)4 ⇒ 𝑓(−1) = 81 Here, L.H.L=R.H.L=f(-1) Hence, f(x) is continuous at a=-1 Question #3 Determine whether the function is continuous at the given point. Show your solution. 𝑘(𝑥) = 𝑥2 at x=-1 𝑥 3 +1 Answer: This question is taken from the continuity and discontinuity in which we have to check whether a given function is continuous at the given point or not continuous, so we can write the given function below 𝑘(𝑥) = 𝑥2 check it at x=-1 is continuous or not 𝑥 3 +1 First, we are written the given function and check whether it is continuous at a given point or not, so we have to check the left hand and right-hand limit (LHL and RHL)are equal or not, Given a function 𝑘(𝑥) = 𝑥2 𝑥 3 +1 so formula we write it below at the given point 𝐿𝐻𝐿 = lim− 𝑘(𝑥) 𝑥→−1 𝑅𝐻𝐿 = lim+ 𝑘(𝑥) 𝑥→−1 and applicable only limits are exist otherwise discontinuous RHL=LHL=k(-1) if the above formula is satisfy then the given function is continuous now we calculate the L.H.L 𝐿𝐻𝐿 = lim− 𝑘(𝑥) 𝑘(𝑥) = 𝑥→−1 𝑥2 𝑥 3 +1 𝐿𝐻𝐿 = lim− 𝑥2 𝑥→−1 𝑥 3 +1 now change the limit x=-1-h and ℎ → 0 𝐿𝐻𝐿 = lim− 𝑥→−1 𝑥2 𝑥 3 +1 (−1−ℎ)2 𝐿𝐻𝐿 = lim (−1−ℎ)3 +1 (1+ℎ2 +2ℎ) ℎ→0 𝐿𝐻𝐿 = lim (−1−ℎ3 ℎ→0 −3ℎ2 −3ℎ+1) put the limits to check indetermine form 1+02 +2×0 𝐿𝐻𝐿 = (−1−03 −3×02 −3×0+1) 1 so the left hand limit does not exist similarly, we check for the RHL 𝑅𝐻𝐿 = lim+ 𝑘(𝑥) 𝑘(𝑥) = 𝑥→1 𝑥2 𝑥 3 +1 𝑅𝐻𝐿 = lim+ 𝑥→−1 𝑥2 𝑥 3 +1 now change the limit x=-1+h and ℎ → 0 1 = (−1+1) = = underfine 0 𝑅𝐻𝐿 = lim+ 𝑥→−1 𝑥2 𝑥3 + 1 (−1+ℎ)2 𝑅𝐻𝐿 = lim (−1+ℎ)3 +1 (1+ℎ2 −2ℎ) ℎ→0 𝑅𝐻𝐿 = lim (−1+ℎ3 ℎ→0 −3ℎ2 +3ℎ+1) put the limits to check indeterminate form (1+02 +2×0) 𝑅𝐻𝐿 = (−1+03 1 −3×02 +2×0+1) 1 = (−1+1) = = undefined 0 Now for the functional value at x=-1 𝑘(𝑥) = 𝑥2 𝑥 3 +1 x=-1 (−1)2 𝑘(−1) = (−1)3 +1 1 1 = (−1)+1 = = undefined 0 So its function value is also not exist as well as LHL and RHL are also undefined so function is discontinuous at x=-1 now we move the next step for the result Result: Result: From the above solution and analysis the functional value, as well as the right and left-hand limits (RHL and LHL ), are also not exist, or at this point, the limits are not defined it means it is not continuous at this point so function 𝑘(𝑥) = 𝑥2 𝑥 3 +1 is at x=-1 is discontinuous or not exist Question #4 Describe the interval on which the function is continuous.  x 2 − 64 , x8  f ( x) =  x − 8 16, x =8  Answer: Consider the function  x 2 − 64 , x8  f ( x) =  x − 8 16, x =8  Check the continuity of the function at x=8. The left hand limit of the function at x=8 is given by lim− 𝑓(𝑥) = lim− ( 𝑥→8 = lim− ( 𝑥→8 = lim− ( 𝑥 2 −82 𝑥→8 𝑥 2 −64 𝑥−8 ) ) 𝑥−8 (𝑥−8)(𝑥+8) 𝑥−8 𝑥→8 ) = lim− (𝑥 + 8) 𝑥→8 =8+8 =16 And the right hand limit at x=8 is lim+ 𝑓(𝑥) = lim+ ( 𝑥→8 2 𝑥→8 2 𝑥 2 −64 𝑥−8 ) 𝑥 −8 ) 𝑥→8 𝑥−8 (𝑥 − 8)(𝑥 + 8) = lim+ ( ) 𝑥→8 𝑥−8 = lim+ (𝑥 + 8) = lim+ ( 𝑥→8 =8+8 =16 Also the function value at x=8 is 16. Since the left hand limit and right hand limits are equal to the function value at x=8. so the limit exist and function is continuous at x=8. Thus, the given function is continuous everywhere, or in interval notation, function is continuous on the interval (−∞, ∞). Question #5 Let f be a function defined below. Discuss the continuity of f at x=-3,x=0, and x=1. If discontinuous, give the type of discontinuity.  x 2 + 3x | x + 3 | x  0, x  −3  f ( x) =  x + 1 if 0  x 1  if x 1  x  Answer: Given The function as  x 2 + 3x | x + 3 | x  0, x  −3  f ( x) =  x + 1 if 0  x 1  if x 1  x  For lim− 𝑓(𝑥) 𝑥→0 f(x)=x+1 if =0 0 n is even 𝑥→±∞ a=1, n=4 3 = ∞2 Apply Infinity Property: ∞𝑐 = =∞ Hence proved f(x) is continuous in its domain (−∞, ∞) Question #9 Evaluate the following limits, where c and k are constants. lim ( 1 𝑥→2 𝑥−2 − 2 𝑥 2 −2𝑥 ) Answer: To evaluate: lim ( 1 𝑥→2 𝑥−2 − 2 𝑥 2 −2𝑥 ) Solution: 1 Given, lim ( 𝑥→2 𝑥−2 − 2 𝑥 2 −2𝑥 ) On simplifying further we get: lim ( 1 𝑥→2 𝑥−2 ⇒ lim ( 2 − ) = lim ( 𝑥 2 −2𝑥 1 2 − 𝑥→2 𝑥−2 1 ⇒ lim ( − ⇒ lim ( − 𝑥→2 𝑥−2 1 𝑥→2 𝑥−2 𝑥 2 −2𝑥 2 𝑥 2 −2𝑥 2 𝑥 2 −2𝑥 1 𝑥→2 𝑥−2 ) = lim ( − 2 ) 𝑥(𝑥−2) 𝑥−2 𝑥→2 𝑥(𝑥−2) 1 ) ) = lim )= 𝑥→2 𝑥 1 2 Result: lim ( 1 𝑥→2 𝑥−2 − 2 𝑥 2 −2𝑥 )= 1 2 Question #10 Find the limits. c and d are attached. circle the answer. 2−𝑦 lim 2 𝑥→∞ √7+6𝑦 Answer: Given lim 2−𝑦 𝑥→∞ √7+6𝑦 2 Solution: Dividing numerator and denominator by y 2−𝑦 lim 2 𝑥→∞ √7+6𝑦 2 −1 𝑦 𝑥→∞ √ 7 +6 𝑦2 lim = 0−1 √0+6 1 =− √6 or − √6 6 Answer: − or − 1 √6 √6 6 Question #11 Evaluate the following limits, where c and k are constants. lim(5𝑥 − 6)3/2 𝑥→2 Answer: 3 To evaluate: lim(5𝑥 − 6)2 𝑥→2 Explanation: 3 lim(5𝑥 − 6)2 𝑥→2 Plug 2 in place of x 3 = (10 − 6)2 3 = (4)2 1 3 = ((4)2 ) 3 = (√4) = (2)3 =8 Question #12 Evaluate the following limits, where c and k are constants. lim (5+ℎ)2 −25 ℎ ℎ→0 Answer: The given limit to be evaluated is lim (5+ℎ)2 −25 ℎ→0 ℎ The limit is evaluated as follows. lim (5+ℎ)2 −25 ℎ→0 = lim ℎ→0 = lim = lim 52 +ℎ2 +10ℎ−25 ℎ ℎ→0 25+ℎ2 +10ℎ−25 ℎ ℎ ℎ(ℎ+10) ℎ ℎ→0 = lim (ℎ + 10) ℎ→0 =10 Question #13 Calculate the following limits, if they exist, by using a combination of polar coordinates and de L’Hopital rule. 1−cos(𝑥 2 +𝑦 2 ) lim 𝑥 2 +𝑦 2 (𝑥,𝑦)→(0,0) Answer: Given: The function is 1−cos(𝑥 2 +𝑦 2 ) lim 𝑥 2 +𝑦 2 (𝑥,𝑦)→(0,0) Consider 𝑥 = 𝑟 cos 𝜃, and 𝑦 = 𝑟 sin 𝜃. The expression becomes lim 1−cos(𝑟 2 ) 𝑟→0 Conclusion: Apply L’Hopital rule as follows: lim 1−cos(𝑟 2 ) 𝑟→0 𝑟2 = lim = lim sin(𝑟 2 ) 𝑟→0 −(− sin(𝑟 2 )⋅2𝑟) 2𝑟 𝑟→0 = sin(0) =0 Hence, lim (𝑥,𝑦)→(0,0) 1−cos(𝑥 2 +𝑦 2 ) 𝑥 2 +𝑦 2 =0 𝑟2 Question #14 Use algebra and limit laws to find the following limits. 5𝑥−5 lim 𝑥→1 3𝑥 2 −𝑥−2 Answer: Given limit, 5𝑥−5 lim 𝑥→1+ 3𝑥 2 −𝑥−2 we have to find the limit using algebraic method. 5𝑥−5 Now, lim+ 2 𝑥→1 3𝑥 −𝑥−2 5𝑥−5 = lim+ 𝑥→1 3𝑥 2 −3𝑥+2𝑥−2 5(𝑥−1) = lim+ 𝑥→1 3𝑥(𝑥+1)+2(𝑥−1) 5(𝑥−1) = lim+ (3𝑥+2)(𝑥−1) 𝑥→1 = lim+ = 𝑥→1 5 5 3𝑥 + 2 5 = =1 3(1)+2 5 Hence solving the limit, we have 5𝑥 − 5 lim =1 𝑥→1+ 3𝑥 2 − 𝑥 − 2 Question #15 Discuss the continuity and differentiability of the function 𝑓(𝑥, 𝑦) = Answer: Given that 𝑓(𝑥, 𝑦) = 𝑥2𝑦 𝑥 2 +𝑦 2 To find the continuity, lim (𝑥,𝑦)→(0,0) = = 𝑓(𝑥, 𝑦) = 𝑥3 lim (𝑥,𝑦)→(0,0) 𝑥 2 +𝑥 2 𝑥2𝑦 lim (𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2 (𝑦 = 𝑥) 𝑥 lim (𝑥,𝑦)→(0,0) 2 =0 Hence 𝑓(𝑥, 𝑦) = 𝑥2𝑦 𝑥 2 +𝑦 2 is continuous at (0,0) Now to find the derivative for f with respect to x, 2𝑥𝑦 𝑓𝑥 = = = 2𝑥𝑦 − 𝑥 2 +𝑦 2 𝑥 2 +𝑦 2 2𝑥𝑦 𝑦2 [1 − 𝑥2 𝑥 2 +𝑦 2 ] 𝑥 2 +𝑦 2 𝑥 2 +𝑦 2 2𝑥𝑦 3 𝑥 2 +𝑦 2 Take the derivative for f with respect to y, 𝑓𝑦 = = = 𝑥2 𝑥 2 +𝑦 2 𝑥2 𝑥 2 𝑦×2𝑦 − (𝑥 2 [1 − 𝑥 2 +𝑦 2 𝑥 2 (𝑥 2 −𝑦 2 ) +𝑦 2 )2 2𝑦 2 𝑥 2 +𝑦 2 ] (𝑥 2 +𝑦 2 )2 The differentiability for f at (0,0) is 𝐷𝑓 = (𝑓𝑥 , 𝑓𝑦 ) = (0,0). Question #16 𝑥2𝑦 𝑥 2 +𝑦 2 at the point (0,0) Calculate the following limits, if they exist, by using a combination of polar coordinates and de L’Hopital rule. arctan(𝑥 2 +𝑦 2 ) lim 𝑥 2 +𝑦 2 (𝑥,𝑦)→(0,0) Answer: changing into polar coordinates 𝑥 = 𝑟 cos 𝜃. 𝑦 = 𝑟 sin 𝜃. 𝑥 2 + 𝑦 2 = 𝑟 2 cos 2 𝜃 + 𝑟 2 sin2 𝜃. = 𝑟 2 (cos 2 𝜃 + sin2 𝜃) = 𝑟 2 (cos 2 𝜃 + sin2 𝜃 = 1) when (𝑥, 𝑦) → (0,0) 𝑟→0 lim Now = lim 𝑥 2 +𝑦 2 (𝑥,𝑦)→(0,0) arctan(𝑟 2 ) 𝑟2 𝑟→0 = lim arctan(𝑥 2 +𝑦 2 ) 1 2 1+(𝑟2 ) ⋅2𝑟 (applying L'Hopital rule) 2𝑟 𝑟→0 1 = lim 𝑟→0 1+𝑟 4 =1 Question #17 Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or −∞, and describe the corresponding vertical asymptote. lim± 𝑥 5 +𝑥−2 2 𝑥→1 𝑥 +𝑥−2 Answer: Given lim 𝑥 5 +𝑥−2 𝑥→1 𝑥 2 +𝑥−2 To find: One sided limit. Vertical asymptotes. Given that lim 𝑥 5 +𝑥−2 𝑥→1 𝑥 2 +𝑥−2 5 Now the equation 𝑥 + 𝑥 − 2 can be factored as (𝑥 − 1)(𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 2) and the equation 𝑥 2 + 𝑥 − 2 can be factored as (𝑥 + 2)(𝑥 − 1). So, lim 𝑥 5 +𝑥−2 𝑥→1 𝑥 2 +𝑥−2 = lim 𝑥→1 (𝑥−1)(𝑥 4 +𝑥 3 +𝑥 2 +𝑥+2) (𝑥−1)(𝑥+2) Cancel out the term (x-1) from numerator and denominator, (𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 2) 𝑥5 + 𝑥 − 2 lim 2 = lim 𝑥→1 𝑥 + 𝑥 − 2 𝑥→1 (𝑥 + 2) = 14 +13 +12 +1+2 = 6 1+2 3 =2 Therefore, 𝑥5 + 𝑥 − 2 lim 2 =2 𝑥→1 𝑥 + 𝑥 − 2 Now to find vertical asymptotes, The line x=a is called vertical asymptote of the curve y=f(x) if lim 𝑓(𝑥) = ±∞ 𝑥→𝑎 For x=-2 (−2)5 + (−2) − 2 𝑥5 + 𝑥 − 2 lim 2 = lim 𝑥→−2 𝑥 + 𝑥 − 2 𝑥→−2 (−2)2 + (−2) − 2 = lim −32−4 𝑥→−2 4−2−2 =∞ Therefore x=-2 is vertical asymptote Therefore, 𝑥5 + 𝑥 − 2 lim 2 =2 𝑥→1 𝑥 + 𝑥 − 2 The vertical asymptote is at x=-2 Question #18 Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or −∞, and describe the corresponding vertical asymptote. lim ± 𝑥2 2 𝑥→−3 𝑥 −9 Answer: The one sided limits are calculated numerically for the given limit. The limit provided in the question is So right hand limit is lim + 𝑥2 𝑥→−3 𝑥 2 −9 = lim + 𝑥2 lim 𝑥2 2 𝑥→−3± 𝑥 −9 𝑥→−3 𝑥 2 −9 9 0+ = +∞ And left hand limit is lim 𝑥2 𝑥→−3− 𝑥 2 −9 = lim − 𝑥2 𝑥→−3 𝑥 2 −9 9 0− = −∞ The vertical asymptote is given by the zeroes of the denominator of any function. So the zeroes of the 2 𝑥2 𝑥 2 −9 is given by, 𝑥 −9 =0 𝑥2 = 9 𝑥 = ±3 Hence, the right hand limit is ∞ and left hand limit is −∞ and the vertical asymptote 𝑥 = ±3 . Question #19 Discuss the continuity of the function  x 2 − 3 x if x  2 f ( x) =  2 x + 4 if x  2 on the open interval 0