Continuity of ? sin ?x and cos ?x a. Use the identi?ty si?n (a ? + ?? )=sin? ?a cos ? ? + co?? sin ?h with the fact that to prove that thereby establishing that ? sin ?x is continuo?us for a ? ll? . ?? in?t: L? ? = ?x ?? ?a? tha? t? = ?a + h? and note that ? h ? ? 0 as ? ? a.) b. Use th? e id?entity co ? s ? +?? ) = co? s ?a ? s ?h ? sin? sin h? with the fact that to prove that .

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 if, 0 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. lx x(x) = f( x ).0 0 Left continuous : lim f() = f(a) , then f(x) is called a left continuous at x=a. xa Right continuous : lim f(x) = f(a) , then f(x) is called a right continuous at x=a. xa+ If , lim f(x) = f(a) = lim f(x) , then f(x) is called a continuous function at x=a. xa xa + If , f(x) is not continuous at x =a means , it is discontinuous at x=a. Step-2 a). Given that ; sin(a+h) = sin(a)cos(h) +cos(a)sin(h)..............(1) and limsin(x) = 0………(2) x0 Now , we have to prove that limsin(x) = sin(a) , by using the conditions (1) and(2). xa Given hint is ; Let h = x a so that x = a + h and note that h 0 as x a. limsin(x) = limsin(x-a+a) , since add and subtract a to x. xa xa Let , take h = x-a , That is , if x = a+h , then as x approaches to a , the value of h approaches to zero.[ since , a=a+h a-a = h 0 = h]. That is , xasin(x) = lxain(x-a+a) = lim sin(h +a) h0 = lim [sin(h)cos(a)+cos(h)sin(a)] , from (1). h0 = cos(a) lim sin(h) +sin(a)lim cos(h),since limCf(x) = Climf(x), h0 h0 xa xa here C is constant function. = cos(a) sin(0) +sin(a) cos(0), take the limits. = cos(a) (0) + sin(a) (1) , since sin(0) =0,and cos(0)=1. = sin(a). Therefore , limsin(x) = sin(a). xa Hence , sin(x) is continuous for all values of x. Step_3 NOTE: sin()is continuous for all of x. 1 1 If = (x) , for all x/ 0. Then g(x) = sin( )x 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: lim(-1) lim sin(1/x) lim(1) x0 x0 x0 -1 x0 sin(1/x) 1,since lxa = C, where C is constant function. That is, the upper limit of x0 sin(1/x) = -1, and the lower limit of lx0sin(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; Step_4 b).Given that ; cos(a+h) = cos(a)cos(h) -sin(a)sin(h)..............(1) and limcos(x) = 1………(2) x0 Now , we have to prove that lxaos(x) = cos(a) , by using the conditions (1) and(2). Given hint is ; Let h = x a so that x = a + h and note that h 0 as x a. limcos(x) = limcos(x-a+a) , since add and subtract a to x. xa xa Let , take h = x-a , That is , if x = a+h , then as x approaches to a , the value of h approaches to zero.[ since , a=a+h a-a = h 0 = h]. That is , xamcos(x) = xacos(x-a+a) = h0 cos(h +a) = lim [cos(h)cos(a)+sin(h)sin(a)] , from(1) h0 = cos(a) lim cos(h) +sin(a)lim sin(h),since limCf(x) = Climf(x), h0 h0 xa xa here C is constant function. = cos(a) cos(0) +sin(a) sin(0), take the limits. = cos(a) (1) + sin(a) (0) , since sin(0) =0,and cos(0)=1. = cos(a). Therefore , limcos(x) = cos(a). xa Hence , cos(x) is continuous for all values of x.