Question

17. imC2x-1x→0 cosx-1, lim

Answer 1

Let the function be,

f( x )= cos2x−1 cosx−1

We have to find the value of the function at limit x→0 .

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

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

f( x )= cos( 2⋅0 )−1 cos( 0 )−1 = cos0−1 cos0−1 = 1−1 1−1 = 0 0

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

Now, we have to simplify the given function.

We know the formula of cos2x and cosx from the trigonometric identities which is:

cos2x=( 1−2 sin 2 x ) cosx=( 1− sin 2 x 2 ) (1)

Using these identities from equation 1, we can now simplify the function;

f( x )= ( 1−2 sin 2 x−1 ) ( 1−2 sin 2 x 2 −1 ) =( −2 sin 2 x −2 sin 2 x 2 ) =( sin 2 x sin 2 x 2 ) (2)

According to the trigonometric theorem,

lim x→0 sinx x =1 (3)

From the theorem of limits, we know that for any two functions f and g , such that both lim x→a f( x ) and lim x→a g( x ) exist, then

lim x→a f( x ) g( x ) = lim x→a f( x ) lim x→a g( x ) (4)

From equations 2 and 4, we get:

lim x→0 sin 2 x sin 2 x 2 = lim x→0 sin 2 x lim x→0 sin 2 x 2 (5)(Applying limits to num and den)

Now we have to solve both numerator and denominator separately and then combine the fraction form.

For solving numerator part, we have to multiply and divide the term with the value x 2 , that is:

lim x→0 sin 2 x x 2 ⋅ x 2 (6)

For solving denominator part we have to multiply and divide the term with the value ( x 2 ) 2 , that is:

lim x→0 sin 2 x 2 ( x 2 ) 2 ⋅ ( x 2 ) 2 (7)

As x→0 , so x 2 →0 .

Combing equations 6, 7 and 8, we get

lim x→0 sin 2 x sin 2 x 2 = lim x→0 sin 2 x lim x→0 sin 2 x 2 = lim x→0 sin 2 x x 2 ⋅ x 2 1 4 lim x→0 sin 2 x 2 ( x 2 ) 2 ⋅ ( x 2 ) 2 =4 lim x→0 sin 2 x x 2 lim x 2 →0 sin 2 x 2 ( x 2 ) 2 =4 lim x→0 ( sinx x ) 2 lim x 2 →0 ( sin x 2 x 2 ) 2

On applying limits and using equation 3, we get:

lim x→0 sin 2 x sin 2 x 2 =4⋅ 1 1 = 4 1

Thus, the value of given lim x→0 sin 2 x sin 2 x 2 =4 .