AE Do removable discontinuities exist? a. Does the fu ? n?cti? ?? ) = ? ? sin (? ) have a removable discontin? uity at ?x =0? b. Does th ? e? function? ?? )= sin (1?/x)have a removable disco ? ntinuity at ? =0?

Solution 87 AE STEP_BY_STEP SOLUTION Step-1 A continuous function can be formally defined as a f unction f : x y ,where the preimage of every open set in y is open in x. More concretely, a function f(x) in a single variable x is said to be continuous at point x i0, 1. If f(x0) is defined, so that x 0 is in the domain of ‘ f’. 2. lim f(x) exists for x in the domain of f. x x0 3. lim f(x) = f( x ). x x0 0 Left continuous : lixa() = f(a) , then f(x) is called a left continuous at x=a. Right continuous : lim f(x+ = f(a) , then f(x) is called a right continuous at x=a. xa If ,xamf(x) = f(a) = lxa+(x) , then f(x) is called a continuous function at x=a. If , f(x) is not continuous at x =a means , it is discontinuous at x=a. Step-2 Removable discontinuity: A hole in a graph. That is, a discontinuity that can be "repaired" by filling in a single point. In other words, a removable discontinuity is a point at which a graph is not connected but can be made connected by filling in a single point. Formally, a removable discontinuity is one at which the limit of the function exists but does not equal the value of the function at that point; this may be because the function does not exist at that point. Step_3 a). The given function is : f(x) = x sin( ) x Now, we need to show that the given function is removable discontinuity at x=0 or not. Now, sin() is a trigonometric function and it is continuous for all values of x, and also this value is lies between [-1, 1]. -1 sin() 1 1 Let , = x then sin(1/x) also lies between [-1,1]. That is , -1 sin(1/x) 1. Now, for x= / 0 multiplying each term of the above equality by x we get -x xsin(1/x) x. Now , take the limits each term of the above equality , then the resultant inequality is: x0(-x) lix0sin(1/x) limx0 0 limx sin(1/x) 0. x0 That is, the upper limit of limx sin(1/x) is zero , and the lower limit of limx sin(1/x) is x0 x0 zero. So, by the squeeze theorem limx sin(1/x) =0 . x0 Therefore , limx sin(1/x) =0 . x0 NOTE: Squeeze theorem : Suppose we have an inequality of functions g(x) f(x) h(x) in an interval around c. Thenlixc(x) lixc(x) lixcx) provided those limits exist. When the limits on the upper bound and lower bound are the same, then the function in the middle is “squeezed” into having the same limit. That is, lim g(x) = L = limh(x). xc xc Then limf(x) = L. xc At , x=0 then the functional value is ; f(0) = (0) sin(1/0) =0 , since -1 sin() 1. Hence the function f(x) = x sin(1/x) have a removable discontinuity at x=0. Hence the graph of f(x) = xsin(1/x) is; Step_4 1 b). The given function is : g(x) = sin( ) x Now, we need to show that the given function is removable discontinuity at x=0 or not. Now, sin() is a trigonometric function and it is continuous for all values of x, and also this value is lies between [-1, 1]. -1 sin() 1 1 Let , = x then sin(1/x) also lies between [-1,1]. That is , -1 sin(1/x) 1. Now , take the limits each term of the above equality , then the resultant inequality is: x0(-1) lx0sin(1/x) lx01) -1 x0 sin(1/x) 1,since xaC = C, where C is constant function. That is, the upper limit ofx0m sin(1/x) = -1, and the lower limit ox0im sin(1/x) =1, are both different we cannot define the value of lim sin(1/x). x0 Therefore , the function f(x) = sin(1/x) does not have a removable discontinuity at x=0. Hence the graph of f(x) = sin(1/x) is;