Skip to content
Miscellaneous 7 II · Q130

Q.lim⁡x→π/4[(sin⁡x−cos⁡x)22−sin⁡x−cos⁡x]\displaystyle\lim_{x\to \pi/4}\left[\frac{(\sin x-\cos x)^2}{\sqrt2-\sin x-\cos x}\right]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
23% · 32/137 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 →

Let t=x−π/4t=x-\pi/4, t→0t\to0. (sin⁡x−cos⁡x)2=1−sin⁡2x(\sin x-\cos x)^2=1-\sin2x; since 2x=π/2+2t2x=\pi/2+2t, sin⁡2x=cos⁡2t\sin2x=\cos2t, so the numerator is 1−cos⁡2t=2sin⁡2t1-\cos2t=2\sin^2t. Also sin⁡x+cos⁡x=2cos⁡t\sin x+\cos x=\sqrt2\cos t (as in Exercise 7.5 II Q2), so the denominator 2−sin⁡x−cos⁡x=2(1−cos⁡t)=22sin⁡2(t/2)\sqrt2-\sin x-\cos x=\sqrt2(1-\cos t)=2\sqrt2\sin^2(t/2). The ratio is 2sin⁡2t22sin⁡2(t/2)=sin⁡2t2sin⁡2(t/2)\dfrac{2\sin^2t}{2\sqrt2\sin^2(t/2)}=\dfrac{\sin^2t}{\sqrt2\sin^2(t/2)}. Since $ …

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.