Question

Differentiate the function with respect to 'x'
${ \pink{ \huge{{{ \lim} \atop{ x\rightarrow \frac{ \pi}{4} }} }}}{ \large{ \frac{ {tan}^{3} x - tan \: x}{cos(x + \frac{x}{4} )} }}$
​

Answer 1

Answer:

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

=Limit x tends to π/4

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

Now applying the limit

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

=2(3–1)/-1 =4/-1 = -4.

Answer 2

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