Question 1 :
Evaluate the following limit
lim x -> 0 √(x2 + a2) - a / √(x2 + b2) - b
Solution :
= lim x -> 0 √(x2 + a2) - a / √(x2 + b2) - b
We may distribute the limits for the numerator and denominator.
= lim x -> 0 √(x2 + a2) - a / lim x -> 0 √(x2 + b2) - b
= lim x->0 [x2+a2-a2/√(x2+a2)+a)]/[x2+b2-b2/√(x2+b2)+b)]
= lim x->0 [x2/√(x2+a2)+a)]/[x2/√(x2+b2)+b)]
= lim x->0 [x2/√(x2+a2)+a)] ⋅ [√(x2+b2)+b)/x2]
= lim x->0 [√(x2+b2)+b)/√(x2+a2)+a)]
= 2b/2a
= b/a
Hence the value of lim x -> 0 √(x2 + a2) - a / √(x2 + b2) - b is b/a.
Question 2 :
Evaluate the following limit
lim x-> 0 (2 arc sinx/3x)
Solution :
= lim x-> 0 (2 sin-1x/3x)
= (2/3) lim x-> 0 (sin-1x/x)
= 2/3
Hence the value of lim x-> 0 (2 arc sinx/3x) is 2/3.
Question 3 :
Evaluate the following limit
lim x-> 0 (1 – cos x)/x2
Solution :
= lim x-> 0 (2 sin2(x/2)/(x2(4/4))
= lim x-> 0 (2/4) sin2(x/2)/(x/2)2
= lim x-> 0 (2/4) (sin(x/2)/(x/2))2
= (1/2) lim x-> 0 (sin(x/2)/(x/2))2
= 1/2
Hence the value of lim x-> 0 (1 – cos x)/x2 is 1/2.
Question 4 :
Evaluate the following limit
lim x-> 0 (tan 2x/x)
Solution :
= lim x-> 0 (tan 2x/x(2/2))
= lim x-> 0 2 (tan 2x/2x)
= 2 lim x-> 0 (tan 2x/2x)
= 2
Hence the value of lim x-> 0 (tan 2x/x) is 2.
Question 5 :
Evaluate the following limit
lim x-> 0 (2x – 3x)/x
Solution :
= lim x-> 0 (2x – 1 + 1 – 3x)/x
= lim x-> 0 [(2x – 1) – (3x - 1)]/x
= [lim x-> 0 (2x – 1)/x] – [lim x-> 0 (3x - 1)/x]
= log 2 – log 3
= log (2/3)
Question 6 :
Evaluate the following limit
lim x-> 0 (3x – 1)/√(x+1) – 1
Solution :
= lim x-> 0 ((3x – 1)/√(x+1) – 1) (√(x+1) + 1 / √(x+1) + 1)
= lim x-> 0 ((3x – 1)/((x+1) – 1) (√(x+1) + 1)
= lim x-> 0 ((3x – 1)/x) (√(x+1) + 1)
= lim x-> 0 ((3x – 1)/x) lim x-> 0 (√(x+1) + 1)
= log 3 (2)
= 2 log 3
= log 32
= log 9
Hence the value of lim x-> 0 (3x – 1)/√(x+1) – 1 is log 9.
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