Limits and continuity #4 - Questions and Answers

Question 1 Find the limit, if it exists, or show that the limits does not exist. lim (x,y)→(3π,2π) xycos(x - 2y) Answer Find the limit: ⇒ ay - ab = bx + ab ⇒ bx + ay = 2ab bx/ab + ay/ab = 2ab/ab ⇒ x/a + y/b = 2 lim (x,y)→(3π,2π) xycos(x - 2y) = (3π)(2π)cos(3π - 2(2π)) = 6π²cos(-π) = 6π²(-1) = -6π² Question 2 Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a. f(x) = (x + 2x³)^4 , a = -1 Answer Limits Limit of the function at point a exists if and only if left hand limit and right hand limit are equal and exists finitely at point a. Left hand limit lim x→a- f(x) = lim h→0 f(a - h) Right hand limit lim x→a+ f(x) = lim h→0 f(a + h) So, limit exists at pont a if and only if lim x→a- f(x) = lim x→a+ f(x) = finite Continuity A function is said to be continuous at a if and only if Limit exits that is lim x→a- f(x) = lim x→a+ f(x) = finite Limit is equal to value of the function at point a that is lim x→a- f(x) = lim x→a+ f(x) = f(a) Solution: We have given function f(x) = (x + 2x³)^4 at points a=-1 Left hand limit lim x→-1- f(x) = lim x→-1- (x + 2x³)^4 = lim h→0 ((-1 - h) + 2(-1 - h)³)^4 = (-1 + 2(-1)³)^4 = (-1 - 2)^4 = (-3)^4 = 81 Hence lim x→-1- f(x) = 81 So limit exists Now, f(-1) = (-1 + 2(-1)³)^4 = (-1 - 2)^4 = (-3)^4 = 81 Therefore, Given function is continuous at x=-1 Question 3 Determine whether the function is continuous at the given point. Show your solution. k(x) = x² / (x³ + 1) at x=-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 / (x³ + 1) check it at x=-1 is continuous or not 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 / (x³ + 1) so formula we write it below at the given point LHL= lim x→-1- k(x) RHL = lim x→-1+ k(x) 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 x→-1- k(x) k(x) = x² / (x³ + 1) LHL = lim x→-1- x² / (x³ + 1) now change the limit x=-1-h and h → 0 LHL = lim x→-1- x² / (x³ + 1) LHL = lim h→0 (-1 - h)² / ((-1 - h)³ + 1) LHL = lim h→0 (1 + h² + 2h) / (-1 - h³ - 3h² - 3h + 1) put the limits to check indetermine form LHL = (1 + 0² + 2 * 0) / (-1 - 0³ - 3 * 0² - 3 * 0 + 1) = 1/0 = undefined so the left hand limit does not exist similarly, we check for the RHL RHL = lim x→-1+ k(x) k(x) = x² / (x³ + 1) RHL = lim x→-1+ x² / (x³ + 1) now change the limit x=-1+h and h → 0 RHL = lim x→-1+ x² / (x³ + 1) RHL = lim h→0 (-1 + h)² / ((-1 + h)³ + 1) Question 4 Use algebra and limit laws to find the following limits. lim x→1+ (5x - 5) / (3x² - x - 2) Answer Given limit, lim x→1+ (5x - 5) / (3x² - x - 2) we have to find the limit using algebric method. Now, lim x→1+ (5x - 5) / (3x² - x - 2) = lim x→1+ (5x - 5) / (3x² - 3x + 2x - 2) = lim x→1+ 5(x - 1) / (3x(x - 1) + 2(x - 1)) = lim x→1+ 5(x - 1) / ((3x + 2)(x - 1)) = lim x→1+ 5 / (3x + 2) = 5 / (3(1) + 2) = 5 / 5 = 1 Hence solving the limit, we have lim x→1+ (5x - 5) / (3x² - x - 2) = 1 Question 5 Limits of composite functions Evaluate each limit and justify your answer lim t→4 tan (t - 4) / (√t - 2) Answer Given limit is: lim t→4 tan ((t - 4) / (√t - 2)) Then we get, lim t→4 tan((t - 4) / (√t - 2)) * (√t + 2) / (√t + 2) lim t→4 tan((t - 4)(√t + 2) / (t - 4)) lim t→4 tan(√t + 2) = tan(√4 + 2) = tan(2+2) = tan 4 Hence the value of this limit is tan 4