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Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is equal to. a) 1 b) -1 c) (sinx)/x d) x/(sinx) The answer is c. Please solve this qn and explain it.?
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Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is e...
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Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is e...
Solution:
1. Breaking down the expression:
We are given the expression cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n) and we need to find its limit as x tends to infinity.
2. Using trigonometric identities:
We can rewrite the expression as cos(x/2) * cos(x/4) * cos(x/8) * ... * cos(x/2^n). This can be further simplified using the identity cos(A)cos(B) = (1/2)[cos(A-B) + cos(A+B)].
3. Applying the identity:
By applying the identity repeatedly, we get the expression as (1/2^n) * [cos(x/2^n) + cos(3x/2^n) + cos(7x/2^n) + ... + cos((2^n-1)x/2^n)].
4. Observing the terms:
As x tends to infinity, the terms inside the brackets oscillate rapidly between -1 and 1. The average value of these terms over a period approaches zero.
5. Limit as x tends to infinity:
Therefore, the limit of the expression as x tends to infinity is 0. However, we need to consider the factor of (1/2^n) which approaches 0 as well.
6. Final result:
Hence, the limit of cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n) as x tends to infinity is 0.
Therefore, the correct answer is a) 1.
1. Breaking down the expression:
We are given the expression cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n) and we need to find its limit as x tends to infinity.
2. Using trigonometric identities:
We can rewrite the expression as cos(x/2) * cos(x/4) * cos(x/8) * ... * cos(x/2^n). This can be further simplified using the identity cos(A)cos(B) = (1/2)[cos(A-B) + cos(A+B)].
3. Applying the identity:
By applying the identity repeatedly, we get the expression as (1/2^n) * [cos(x/2^n) + cos(3x/2^n) + cos(7x/2^n) + ... + cos((2^n-1)x/2^n)].
4. Observing the terms:
As x tends to infinity, the terms inside the brackets oscillate rapidly between -1 and 1. The average value of these terms over a period approaches zero.
5. Limit as x tends to infinity:
Therefore, the limit of the expression as x tends to infinity is 0. However, we need to consider the factor of (1/2^n) which approaches 0 as well.
6. Final result:
Hence, the limit of cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n) as x tends to infinity is 0.
Therefore, the correct answer is a) 1.
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Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is equal to. a) 1 b) -1 c) (sinx)/x d) x/(sinx) The answer is c. Please solve this qn and explain it.?
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Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is equal to. a) 1 b) -1 c) (sinx)/x d) x/(sinx) The answer is c. Please solve this qn and explain it.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is equal to. a) 1 b) -1 c) (sinx)/x d) x/(sinx) The answer is c. Please solve this qn and explain it.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is equal to. a) 1 b) -1 c) (sinx)/x d) x/(sinx) The answer is c. Please solve this qn and explain it.?.
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Solutions for Limit x tends to infinite cos(x/2)cos(x/2^2)cos(x/2^3).cos(x/2^n) is equal to. a) 1 b) -1 c) (sinx)/x d) x/(sinx) The answer is c. Please solve this qn and explain it.? in English & in Hindi are available as part of our courses for JEE.
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