x→4πlim2−2sin2x82−(cosx+sinx)7(00form)
=x→4πlim−22cos2x−7(cosx+sinx)6(−sinx+cosx) (using L'Hospital Rule)
=x→4πlim22cos2x56(cosx−sinx)(00form)
=x→4πlim−42sin2x−56(sinx+cosx) (using L'Hospital Rule)
=42562=14
JEE Main 2022 — Mathematics Calculus
x→4πlim2−2sin2x82−(cosx+sinx)7 is equal to
Held on 25 Jul 2022 · Verified 6 Jul 2026.
14
7
142
72
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