We have,

lim

x→0

cos2x−1

cosx−1

=

lim

x→0

cos2x−1

−(−1−cosx)

=

lim

x→0

(1−cos2x)

(1−cosx)

×

(1+cosx)

(1+cosx)

=

lim

x→0

(2sin2x)(1+cosx)

1−cos2x

=

lim

x→0

2(1+cosx)
= 2 (1 + cos 0) = 2 (1 + 1) = 2 × 2 = 4.
Alternative solution :

lim

x→0

cos2x−1

cosx−1

=

lim

x→0

1−2sin2x−1

1−2sin2 x 2 −1

=

lim

x→0

sin2x

sin2 x 2

=

lim

x→0

sin2x

(x2)

×x2×

( x 2 )2

sin2 x 2

×

1

( x 2 )2

=

lim

x→0

sin2x

x2

.

lim

x→0

( x 2 )

sin2 x 2

×x2×

4

x2

= 4