Limits Continuity and Differentiability

Given, function
f(x)={

‌ (eax−1)‌log(1+x) sin‌2x	,‌	x>0
‌ cos‌4‌x−cos‌b‌x tan2x	,‌	x=0

is contiuous at x=0
Since, f(x) is continuous at x=0
So,

lim

x⟶0−

f(x)=

lim

x⟶0+

f(x)=f(0) and f(0)=2
Now,

lim

x⟶0+

‌

(eax−1)‌log(1+x)

sin‌2x

Applying L'hospital rule, twice, since it is an indeterminate form ‌

0

0

.
So,

lim

x⟶0+

‌

aeax‌log(1+x)+ (Eax−1) 1+x

2‌cos‌x‌s‌i‌n‌x

This limit is again ‌

0

0

form, so apply L'Hospital rule again.
‌

lim

x⟶0+

‌

a2eax‌log(1+x)+ aeax(1+x)−(eax−1) (1+x)2 + aeax 1+x

2cos2x−2sin‌2x

‌=‌

0+ a(1)−(1−1) 12 + a 1

2(1)−0

=‌

a+a

2

=a
Since,

lim

x⟶0+

f(x)=f(0)⇒a=2
Now,

lim

x⟶0−

‌

cos‌4‌x−cos‌b‌x

tan2x

‌=

lim

x⟶0−

‌

−2sin‌( 4x+bx 2 )sin‌( 4x−bx 2 )

tan2x

‌=

lim

x⟶0−

‌

−2sin‌( 4x+bx 2 )sin‌( 4x−bx 2 )⋅cos2x

sin‌2x x2 ⋅x2

‌

lim

x⟶0−

‌

‌ −2sin‌( 4x+bx 2 ) 4x+bx 2 ⋅( 4x+bx 2 )⋅ sin‌( 4x−bx 2 ) ( 4x−bx 2 ) ( 4x−bx 2 )⋅cos2x

sin‌2x x2 ⋅x2

‌=

lim

x⟶0−

‌

−2⋅1⋅( 4x+bx 2 )⋅1⋅( 4x−bx 2 )⋅cos2x

1⋅x2

‌‌[∵

lim

x⟶0

‌

sin‌x

x

=1]
‌=

lim

x⟶0−

‌

−2(16x2−b2x2)

4x2

⋅cos2x
‌=

lim

x⟶0−

‌

(b2−16)

2

⋅cos2x=‌

b2−16

2

⋅1‌‌[∵

lim

x⟶0−

cos2x=1]
‌=‌

b2−16

2

Also,

lim

x⟶0−

f(x)=f(0)
⇒‌‌‌

b2−16

2

=2⇒b2=4+16=20
Now, √b2−a2=√20−22=√20−4=√16=4