# Calculate Limits of Trigonometric Functions

 Several examples with detailed solutions and exercises with answers are presented. Example 1: Find the limit limx→0 (1 - cos x) / x Solution to Example 1: Let us multiply the numerator and denominator of (1 - cos x) / x by (1 + cos x) and write limx→0 (1 - cos x) / x = limx→0 [ (1 - cos x) / x ] *[ (1 + cos x) / (1 + cos x) ] The numerator becomes 1 - cos 2 x = sin 2 x, hence limx→0 (1 - cos x) / x = limx→0 [ (sin 2 x) / x ] *[ 1/ (1 + cos x) ] The limit can be written limx→0 (1 - cos x) / x = limx→0 [ (sin x) / x ] * limx→0 [ sin x / (1 + cos x) ] = (1)(0/2) = 0 We have used the theorem: limx→0 [ (sin x) / x ] = 1 Example 2: Find the limit limx→0 sin 4 x / 4 x Solution to Example 2: Let t = 4x. When x approaches 0, t = 4x approaches 0, so that limx→0 sin 4 x / 4 x = limt→0 sin t / t We now use the theorem: limt→0 sin t / t = 1 to find the limit Find the limit limx→0 sin 4 x / 4 x = limt→0 sin t / t = 1 Example 3: Find the limit limx→0 sin 6 x / 5 x Solution to Example 3: Let t = 6 x or x = t / 6. When x approaches 0, t = 6 x approaches 0, so that limx→0 sin 6 x / 5 x = limt→0 sin t / (5 t / 6) = limt→0 (6 / 5) sin t / t = (6 / 5) limt→0 sin t / t = (6 / 5) * 1 = 6 / 5 Example 4: Find the limit limx→-3 sin (x + 3) / (x 2 +7x + 12) Solution to Example 4: If we apply the theorem of the limit of the quotient of two functions, we will get the indeterminate form 0 / 0. We need to find another way. For x = -3, the denominator is equal to zero and therefore may be factorized, hence limx→-3 sin (x + 3) / (x 2 +7x + 12) = limx→-3 sin (x + 3) / (x + 3)(x + 4) Let t = x + 3 or x = t - 3. As x approaches -3, t approaches 0. limx→-3 sin (x + 3) / (x 2 +7x + 12) = limx→-3 sin (x + 3) / (x + 3)(x + 4) = limt→0 sin t / [ t (t + 1) ] We now apply the theorem of the limit of the product of two functions. = limt→0 sin t / t * limt→0 1 / (t + 1) = (1)*(1) = 1 Example 5: Find the limit limx→0 sin | x | / x Solution to Example 5: We shall find the limit as x approaches 0 from the left and as x approaches 0 from the right. For x < 0, | x | = -x limx→0 - sin | x | / x = limx→0 - sin (- x ) / x = - limx→0 - sin ( x ) / x = -1 For x > 0, | x | = x limx→0 + sin | x | / x = limx→0 + sin x / x = 1 The two limits from the left and from the right are different, therefore the above limit does not exist. limx→0 sin | x | / x does not exist Example 6: Find the limit limx→0 x / tan x Solution to Example 6: We first use the trigonometric identity tan x = sin x / cos x = -1 limx→0 x / tan x = limx→0 x / (sin x / cos x) = limx→0 x cos x / sin x = limx→0 cos x / (sin x / x) We now use the theorem of the limit of the quotient. = [ limx→0 cos x ] / [ limx→0 sin x / x ] = 1 / 1 = 1 Example 7: Find the limit limx→0 x csc x Solution to Example 7: We first use the trigonometric identity csc x = 1 / sin x limx→0 x csc x = limx→0 x / sin x = limx→0 1 / (sin x / x) The limit of the quotient is used. = 1 / 1 = 1 Exercises: Find the limits 1. limx→0 (sin 3x / sin 8x) 2. limx→0 tan 3x / x 3. limx→0 sqrt(x) csc [ 4sqrt(x) ] 4. limx→0 sin 3 3x / x sin(x 2) Solutions to Above Exercises: Find the limits 1. 3 / 8 2. 3 3. 1 / 4 4. 27 More on limits Calculus Tutorials and Problems