Question 1 :
(1) (a) Find the left and right limits of
f(x) = (x2 - 4) / (x2 + 4x + 4) (x + 3) at x = -2
Solution :
f(x) = (x2 - 4) / (x2 + 4x + 4) (x + 3)
f(x) = (x + 2)(x - 2) / (x + 2)(x + 2)(x + 3)
f(x) = (x - 2)/(x + 2)(x + 3)
= lim x->-2-(x-2)/(x+2)(x +3)
To find left hand limit of -2-, let us take -2.1
= (-2.1-2)/(-2.1+2)(-2.1+3)
= -4.1/(-0.1)(0.9)
= ∞
To find left hand limit of -2+, let us take -1.9
= (-1.9-2)/(-1.9+2)(-1.9+3)
= -3.9/(0.1)(1.1)
= -∞
Hence the left hand limit of -2- is ∞ and right hand limit of -2+ is -∞.
(b) f (x) = tan x at x = π/2
Solution :
f (x) = tan x
f(x) = lim x->π/2- tan x
The angle lesser than 90 degree lies in the first quadrant, for all trigonometric ratios we will get positive sign.
f(x) = lim x->π/2- tan (π/2)
= ∞
The angle greater than 90 degree lies in the second quadrant, for sin and cosec, we will have positive and for tan we will get negative sign.
f(x) = lim x->π/2- tan (π/2)
= -∞
Question 2 :
Evaluate the following
lim x->3 (x2 - 9) / x2(x2 - 6x + 9)
Solution :
= lim x->3 (x2 - 9) / x2(x2 - 6x + 9)
= lim x->3 (x + 3)(x - 3) / x2(x-3)2
= lim x->3 (x + 3) / x2(x-3)
The simplified form does not match with any formulas in limits, so let us find left hand and right hand limit.
Left hand limit : = lim x->3- (x+3)/x2(x-3) = -∞ |
Right hand limit : = lim x->3+ (x+3)/x2(x-3) = ∞ |
Question 3 :
lim x->∞ [3/(x - 2) - (2x + 11)/(x2 + x - 6)]
Solution :
f(x) = 3/(x - 2) - (2x + 11)/(x2 + x - 6)
f(x) = [3/(x - 2) - (2x + 11)/(x-2) (x+3)]
= 3(x+3)-(2x+11)/(x-2) (x+3)
= (3x+9-2x-11)/(x-2) (x+3)
= (x - 2)/(x - 2)(x + 3)
= 1/(x + 3)
= lim x->∞[1/(x + 3)]
Dividing by the highest exponents of denominator, we get
= lim x->∞[(1/x)/(1 + 3/x)]
By applying ∞ instead of x, we get
= 0
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