Question

$\lim_{x \to\pi } sin(\pi cosx)$

Answer 1

Answer:

There’s nothing to do but substitute for x . You can do this because it does not result in an indeterminate form.

L=limx→πsin(πcosx)

⟹L=sin(πcosπ)

⟹L=sin[π⋅(−1)]

⟹L=sin(−π)

⟹L=0

Step-by-step explanation:

I connot send you written solution because I don't have my pen and notebook with me right now,

But let me tell you what to do

First look at that x-pie/3 inside the the argument of sin. How about changing it to a less scarier form?

Step 1 substitute x-pie/3 =t. Please note if x tend to pie/3, t tend to 0

Now write it again like

Lim. (sin(t) /(1–2cos(t+pie/3)). (equation 1 let's say)

t - - - > 0.

Step 2

Now I hope you know the property cos(a+b) =cosa. Cosb-sina. Sinb

Use this property and in equation 1

You get

Lim . (sin(t) /(1–cost+root3. Sint)

t—— - >0

Step 3 you know 1-cosx=2sin^2(x/2)

Apply this you get

Lim . . (sin(t) /(2sin^2(t/2) + root 3. Sint)

t——-> 0

Form now I think u can solve but let me finish this

Use sint/t= 1 now and solve u will get some answer

Answer =1/root3

Answer 2

limₓ --> π sin (πcosx) = 1/√3

GIVEN: π sin (πcosx)

TO FIND: limₓ --> π sin (πcosx)

SOLUTION:
As we know,

To find the limit we shall substitute the value of x.

As we are given in the question,

limₓ→πsin(πcosx)

= sin(πcosπ)

= sin[π⋅(−1)]

= sin(−π)

=0

For limit of x tending to pie,

where sin has pie cos x as its theta.

Step1 : Let us substitute  x-pie/3 =t

Therefore,

Lim. (sin(t) /(1–2cos(t+pie/3)).                                           ----Eq1

where,

t - - - > 0.

Step 2: Using the property cos(a+b) =cosa. Cosb-sina. Sinb in Equation1

Lim . (sin(t) /(1–cost+root3. Sint)

where,

t—— - >0

Step 3: As we know 1-cosx=2sin^2(x/2)

Apply this you get in the above equation,

Lim . . (sin(t) /(2sin^2(t/2) + root 3. Sint)

where,

t——-> 0

Using sin/t = 1 we get,

Answer =1/√3

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