lim θ-->0 (sinθ/θ) = 1
limθ-->0 (1 - cosθ)/θ = 0
limθ-->0 (tanθ)/θ = 1
limx-->0 (sin-1x)/x = 1
limx-->0 (tan-1x)/x = 1
limx-->a [sin(x - a)]/(x - a) = 1
limx-->a [tan(x - a)]/(x - a) = 1
Question 1 :
Evaluate the following limit
lim x -> 0 sin3 (x/2)/ x3
Solution :
lim x -> 0 sin3 (x/2)/ x3 = lim x -> 0 (sin (x/2))3/ x3
In order to match the given question with the formula, let us multiply and divide by 1/8
= lim x -> 0 (sin (x/2))3(1/8)/ x3(1/8)
= (1/8) lim x -> 0 (sin (x/2))3/ (x/2)3
= (1/8) lim x -> 0 [sin (x/2) / (x/2)]3
= 1/8 (1)
= 1/8
Hence the value of lim x -> 0 sin3 (x/2)/ x3 is 1/8.
Question 2 :
Evaluate the following limit
lim x -> 0 sin αx / sin ᵦx
Solution :
= lim x -> 0 sin αx / sin ᵦx
= (lim x -> 0 sin αx) (αx/αx) / (lim x -> 0 sin ᵦx)(ᵦx/ᵦx)
= (αx/ᵦx) (lim x -> 0 sin αx/αx) / (lim x -> 0 sin ᵦx/ᵦx)
= (α/ᵦ) (1)
= α/ᵦ
Hence the value of lim x -> 0 sin αx / sin ᵦx is α/ᵦ.
Question 2 :
Evaluate the following limit
lim x -> 0 tan 2x / sin 5x
Solution :
= lim x -> 0 tan 2x / sin 5x
= lim x -> 0 tan 2x ⋅ (2x/2x) / lim x -> 0 sin 5x ⋅(5x/5x)
= (2x/5x) [(lim x -> 0 tan 2x/2x) / (lim x -> 0 sin 5x/5x)
= (2/5) (1/1)
= 2/5
Hence the value of lim x -> 0 tan 2x / sin 5x is 2/5.
Question 3 :
Evaluate the following limit
lim α -> 0 (sin αn)/ (sin α)m
Solution :
= lim α -> 0 (sin αn) ⋅ (αn/αn) / (sin α ⋅ (α/α))m
= lim α -> 0 (αn/αm) lim α -> 0 (sin αn/αn) /lim α -> 0 (sin α /α)m
= lim α -> 0 (αn/αm)
= lim α -> 0 αn-m
If n = m = lim α -> 0 αn-n = 0n - n = 1 |
If m > n = lim α -> 0 αn-m = 0n - m = 0negative value = 0 |
If m < n = lim α -> 0 αn-m = 0n - m = 0positive value = ∞ |
Question 4 :
Evaluate the following limit
lim x -> 0 (sin (a + x) - sin (a - x))/ x
Solution :
= lim x -> 0 (sin (a + x) - sin (a - x)) / x
sin C - sin D = 2 cos ((C + D)/2) sin ((C - D)/2)
= lim x -> 0 (2 cos ((a+x+a-x)/2) sin (a+x-a+x)/2) / x
= lim x -> 0 (2 cos a sin x)/x
= 2 cos a lim x -> 0 (sin x/x)
= 2 cos a (1)
= 2 cos a
Hence the value lim x -> 0 (sin (a + x) - sin (a - x))/ x is 2 cos a.
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