# HOW TO FIND LEFT AND RIGHT LIMITS

## About "How to Find Left and Right Limits"

How to Find Left and Right Limits :

Here we are going to see 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/2tan 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/2tan (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/2tan (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

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Left and Right Limits"

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