Several examples with detailed solutions and exercises with answers are presented.
Example 1: Find the limit
lim_{x→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
lim_{x→0} (1  cos x) / x
= lim_{x→0} [ (1  cos x) / x ] *[ (1 + cos x) / (1 + cos x) ]
The numerator becomes 1  cos ^{ 2} x = sin ^{ 2} x, hence
lim_{x→0} (1  cos x) / x
= lim_{x→0} [ (sin ^{ 2} x) / x ] *[ 1/ (1 + cos x) ]
The limit can be written
lim_{x→0} (1  cos x) / x
= lim_{x→0} [ (sin x) / x ] * lim_{x→0} [ sin x / (1 + cos x) ] = (1)(0/2) = 0
We have used the theorem: lim_{x→0} [ (sin x) / x ] = 1
Example 2: Find the limit lim_{x→0} sin 4 x / 4 x
Solution to Example 2:
Let t = 4x. When x approaches 0, t = 4x approaches 0, so that
lim_{x→0} sin 4 x / 4 x = lim_{t→0} sin t / t
We now use the theorem: lim_{t→0} sin t / t = 1 to find the limit
Find the limit lim_{x→0} sin 4 x / 4 x = lim_{t→0} sin t / t = 1
Example 3: Find the limit lim_{x→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
lim_{x→0} sin 6 x / 5 x = lim_{t→0} sin t / (5 t / 6)
= lim_{t→0} (6 / 5) sin t / t
= (6 / 5) lim_{t→0} sin t / t
= (6 / 5) * 1 = 6 / 5
Example 4: Find the limit lim_{x→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
lim_{x→3} sin (x + 3) / (x^{ 2} +7x + 12)
= lim_{x→3} sin (x + 3) / (x + 3)(x + 4)
Let t = x + 3 or x = t  3. As x approaches 3, t approaches 0.
lim_{x→3} sin (x + 3) / (x^{ 2} +7x + 12)
= lim_{x→3} sin (x + 3) / (x + 3)(x + 4)
= lim_{t→0} sin t / [ t (t + 1) ]
We now apply the theorem of the limit of the product of two functions.
= lim_{t→0} sin t / t * lim_{t→0} 1 / (t + 1)
= (1)*(1) = 1
Example 5: Find the limit lim_{x→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
lim_{x→0 } sin  x  / x
= lim_{x→0 } sin ( x ) / x
=  lim_{x→0 } sin ( x ) / x
= 1
For x > 0,  x  = x
lim_{x→0 +} sin  x  / x
= lim_{x→0 +} sin x / x
= 1
The two limits from the left and from the right are different, therefore the above limit does not exist.
lim_{x→0} sin  x  / x does not exist
Example 6: Find the limit lim_{x→0} x / tan x
Solution to Example 6:
We first use the trigonometric identity tan x = sin x / cos x
= 1
lim_{x→0} x / tan x
= lim_{x→0} x / (sin x / cos x)
= lim_{x→0} x cos x / sin x
= lim_{x→0} cos x / (sin x / x)
We now use the theorem of the limit of the quotient.
= [ lim_{x→0} cos x ] / [ lim_{x→0} sin x / x ] = 1 / 1 = 1
Example 7: Find the limit lim_{x→0} x csc x
Solution to Example 7:
We first use the trigonometric identity csc x = 1 / sin x
lim_{x→0} x csc x
= lim_{x→0} x / sin x
= lim_{x→0} 1 / (sin x / x)
The limit of the quotient is used.
= 1 / 1 = 1
Exercises: Find the limits
1. lim_{x→0} (sin 3x / sin 8x)
2. lim_{x→0} tan 3x / x
3. lim_{x→0} sqrt(x) csc [ 4sqrt(x) ]
4. lim_{x→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
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