Skip to content
7.4 Q.III · Q58

Q.lim⁡x→0[cos⁡(ax)−cos⁡(bx)cos⁡(cx)−1]\displaystyle\lim_{x\to 0}\left[\frac{\cos(ax)-\cos(bx)}{\cos(cx)-1}\right]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
71% · 97/137 Questions
✓ Free question

cos⁡(ax)−cos⁡(bx)=−2sin⁡ ⁣((a+b)x2)sin⁡ ⁣((a−b)x2)\cos(ax)-\cos(bx)=-2\sin\!\left(\dfrac{(a+b)x}{2}\right)\sin\!\left(\dfrac{(a-b)x}{2}\right), and cos⁡(cx)−1=−2sin⁡2 ⁣(cx2)\cos(cx)-1=-2\sin^2\!\left(\dfrac{cx}{2}\right). As x→0x\to0, each small-angle sine behaves like its argument: sin⁡ ⁣((a+b)x2)∼(a+b)x2\sin\!\left(\tfrac{(a+b)x}2\right)\sim\tfrac{(a+b)x}2, sin⁡ ⁣((a−b)x2)∼(a−b)x2\sin\!\left(\tfrac{(a-b)x}2\right)\sim\tfrac{(a-b)x}2, so the numerator ∼(a+b)(a−b)x24×(−2)×(−1)=(a2−b2)x24×2\sim\tfrac{(a+b)(a-b)x^2}{4}\times(-2)\times(-1)=\tfrac{(a^2-b^2)x^2}{4}\times2... more directly: numerator/x2→(a+b)(a−b)4×...x^2\to\dfrac{(a+b)(a-b)}{4}\times... Carrying the constants through carefully: numerator →−2⋅(a+b)x2⋅(a−b)x2=−(a2−b2)x22\to -2\cdot\tfrac{(a+b)x}2\cdot\tfrac{(a-b)x}2=-\tfrac{(a^2-b^2)x^2}{2}, and denominator →−2(cx2)2=−c2x22\to-2\left(\tfrac{cx}2\right)^2=-\tfrac{c^2x^2}{2}. Dividing: −(a2−b2)x2/2−c2x2/2=a2−b2c2\dfrac{-(a^2-b^2)x^2/2}{-c^2x^2/2}=\dfrac{a^2-b^2}{c^2}.

✓Final answer

a2−b2c2\dfrac{a^2-b^2}{c^2}

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.