$\underset{x→∞}{\lim}\frac{\sin^4x-\sin^2x+1}{\cos^4x-\cos^2x+1}$ is equal to
Answer & explanation
Correct answer: option 2
$\sin^4x-\sin^2x+1$
$=(1-\cos^2x)^2-(1-\cos^2x)+1=1+\cos^4x-\cos^2x⇒\frac{\sin^4x-\sin^2x+1}{\cos^4x-\cos^2x+1}=1$
Thus given limit is equal to one.