HOW TO FIND LEFT AND RIGHT LIMITS

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.

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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