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$\underset{x→∞}{\lim}\frac{\sin^4x-\sin^2x+1}{\cos^4x-\cos^2x+1}$ is equal to |
0 1 1/3 1/2 |
1 |
$\sin^4x-\sin^2x+1$ $=(1-\cos^2x)^2-(1-\cos^2x)+1=1+\cos^4x-\cos^2x⇒\frac{\sin^4x-\sin^2x+1}{\cos^4x-\cos^2x+1}=1$ Thus given limit is equal to one. |
