Question

tan 2x22· 1-1

Answer 1

Let the function be,

f( x )= tan2x ( x− π 2 )

We have to find the value of the function at limit x→ π 2 .

So we need to check the function by substituting the value at particular point (-2), so that it should not be of the form. 0 0

If the condition is true, then we need to simplify the terms to remove 0 0 form.

f( x )= tan( 2⋅ π 2 ) ( π 2 − π 2 ) = tanπ 0 = 0 0

Here, we see that the condition is not true and it is in 0 0 form.

So we need to reduce the function to its simplest and standard form.

The given expression lim x→ π 2 ( x− π 2 ) can be solved by assumption,

Let the value of ( x− π 2 ) be y . (1)

As x→ π 2 , so y→0

And from equation (1), we get x=( y+ π 2 ) . (2)

On substituting the value of new limits in terms of y from equations (1) and (2), we get

lim y→0 tan( 2⋅( y+ π 2 ) ) y = lim y→0 tan( π+2y ) y (3)

From the trigonometric identity we know that:

tan(π+2θ)=tan2θ (4)

Combining equations (3) and (4), we get

lim y→0 tan( π+2y ) y = tan2y y

We need to further simply the expression into more general form to remove 0 0 form.

tan2y y = sin2y cos2y y = sin2y y⋅cos2y (5)

By multiplying and dividing equation (5) with a constant value 2, we get

sin2y y⋅cos2y = sin2y⋅2 2y⋅cos2y =( sin2y 2y )( 2 cos2y ) (6)

Now, limits can be easily applied on equation (6):

lim y→0 ( ( sin2y 2y )( 2 cos2y ) )= lim y→0 ( sin2y 2y )⋅ lim y→0 ( 2 cos2y ) (7)

As y→0 , so 2y→0 .

According to the trigonometric theorem,

lim x→0 sinx x =1 (8)

Solving the limits of equation (7) with the help of equation (8), we get

lim 2y→0 ( sin2y 2y )⋅ lim y→0 ( 2 cos2y )=1⋅( 2 cos2⋅0 ) =1⋅( 2 cos0 ) =2

Thus, the value of the given expression lim x→ π 2 tan2x ( x− π 2 ) =2