Answer:

Lim x tends to π/4. This is of the form 0/0. So by using l’Hopital’s rule we can write it as

Lim x tends to π/4

=Limit x tends to π/4

Now applying the limit

=2(3–1)/-1 =4/-1 = -4.

Answer:

Lim x tends to π/4[(tan^3x- tan x)/{cos (x+π/4)}][(tan

3

x−tanx)/cos(x+π/4)] . This is of the form 0/0. So by using l’Hopital’s rule we can write it as

Lim x tends to π/4[{3tan^2 x.sec^2 x - sec^2 x}/{-sin(x+π/4)}[3tan

2

x.sec

2

x−sec

2

x/−sin(x+π/4)

=Limit x tends to π/4

[{sec^2 x(3tan^2 x - 1)}/{- sin(x+π/4)}[sec

2

x(3tan

2

x−1)/−sin(x+π/4)

Now applying the limit

[sec^2 (π/4){3tan^2 (π/4) - 1}/-sin(π/4+π/4)[sec

2

(π/4)3tan

2

(π/4)−1/−sin(π/4+π/4)

=2(3–1)/-1 =4/-1 = -4