Exams › JEE Main › Maths

Evaluate the limit as x approaches 0 of (8/x⁸) times [1 - cos(x²/2) - cos(x²/4) + cos(x²/2)cos(x²/4)].

1. 1/8
2. 1/16
3. 1/32
4. 1/64

Correct answer: 1/32

Solution

The bracket = (1-cos(x^2/2))(1-cos(x^2/4)). Using 1-cos(u)~u^2/2: 1-cos(x^2/2)~x^4/8 and 1-cos(x^2/4)~x^4/32, so the product ~x^8/256. Multiplying by 8/x^8 gives 8/256 = 1/32.

Related JEE Main Maths questions

• Evaluate the limit as x approaches 0: ((1+tan x)/(1+sin x))^(csc x).
• Evaluate the limit as x approaches 0 of sin([x−3]) divided by [x−3], where [.] denotes the greatest integer function.
• Evaluate the limit as x approaches 0: (sin(x⁴) − x⁴ cos(x⁴) + x²⁰) / (x⁴(e^(2x⁴) − 1 − 2x⁴)).
• Evaluate the limit lim_(n→∞)(n³-8)/(n³+8). What is its value?
• If lim_(x→∞) (1+(λ)/(x)+(μ)/(x²))^(2x)=e², which of the following values of λ and μ is correct?
• Consider the following statements: Statement 1: If lim_(x→α)(sin(f(x)))/(x-α), where f(x)=ax²+bx+c, exists as a finite nonzero number, then lim_(x→α)(1/f(x)-1)/(1/f(x)+1) does not exist. Statement 2: The limit lim_(x→α)(sin(f(x)))/(x-α) can be finite only in the indeterminate form 0/0.

⚔️ Practice JEE Main Maths free + battle 1v1 →