Fsc Part 2 Mathematics (Complete Solutions)

# Q1 Determine the left hand limit and right hand limit and then find limit of the following function

Determine the left hand limit and right hand limit and then find limit of the following function
when x → c
(i)    f(x) = 2x2 + x - 5,    c = 1      (ii)    f(x) =
 x2 - 9 x - 3
,    c = -3
(iii)    f(x) = |x - 5|,    c = 5

Solution:

(i)    f(x) = 2x2 + x - 5,    c = 1

 Left hand limit = Lim x → 1- f(x) = Lim x → 1- (2x2 + x - 5) = 2(1)2 + 1 - 5 = 2 + 1 - 5 = -2

 Right hand limit = Lim x → 1+ f(x) = Lim x → 1+ (2x2 + x - 5) = 2(1)2 + 1 - 5 = 2 + 1 - 5 = -2

 Hence Lim x → 1- f(x) = Lim x → 1+ f(x) = -2

 ∴ Lim x → 1 ƒ(x) = -2

(ii)    f(x) =
 x2 - 9 x - 3
,    c = -3

Left hand limit = Lim
x → -3-
f(x)
= Lim
x → -3-
 x2 - 9 x - 3
=
 (-3)2 - 9 (-3) - 3
=
 9 - 9 -6
=
 0 -6
= 0

Right hand limit = Lim
x → -3+
f(x)
= Lim
x → -3+
 x2 - 9 x - 3
=
 (-3)2 - 9 (-3) - 3
=
 9 - 9 -6
=
 0 -6
= 0

 Hence Lim x → -3- f(x) = Lim x → -3+ f(x) = 0

 ∴ Lim x → -3 ƒ(x) = 0

(iii)    f(x) = |x - 5|,    c = 5

 Left hand limit = Lim x → 5- f(x) = Lim x → 5- |x - 5| = |5 - 5| = |0| = 0

 Right hand limit = Lim x → 5+ f(x) = Lim x → 5+ |x - 5| = |5 - 5| = |0| = 0

 Hence Lim x → 5- f(x) = Lim x → 5+ f(x) = 0

 ∴ Lim x → 5 ƒ(x) = 0