Skip to content
Miscellaneous 7 II · Q136

Q.lim⁡x→0{1x12[1−cos⁡(x22)−cos⁡(x44)+cos⁡(x22)cos⁡(x44)]}\displaystyle\lim_{x\to 0}\left\{\frac{1}{x^{12}}\left[1-\cos\left(\frac{x^2}{2}\right)-\cos\left(\frac{x^4}{4}\right)+\cos\left(\frac{x^2}{2}\right)\cos\left(\frac{x^4}{4}\right)\right]\right\}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
28% · 38/137 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

With A=cos⁡(x2/2)A=\cos(x^2/2) and B=cos⁡(x4/4)B=\cos(x^4/4), the bracket 1−A−B+AB=(1−A)(1−B)1-A-B+AB=(1-A)(1-B). Using the half-angle identity, 1−A=2sin⁡2(x2/4)1-A=2\sin^2(x^2/4) and 1−B=2sin⁡2(x4/8)1-B=2\sin^2(x^4/8). So the bracket =4sin⁡2(x2/4)sin⁡2(x4/8)=4\sin^2(x^2/4)\sin^2(x^4/8). As x→0x\to0: sin⁡(x2/4)∼x2/4\sin(x^2/4)\sim x^2/4 so sin⁡2(x2/4)∼x4/16\sin^2(x^2/4)\sim x^4/16; sin⁡(x4/8)∼x4/8\sin(x^4/8)\sim x^4/8 so sin⁡2(x4/8)∼x8/64\sin^2(x^4/8)\sim x^8/64. The bracket $\sim4 …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.