Question

$\underset{x\to \infty }{lim}\frac{{\sin }^{4}x-{\sin }^{2}x+1}{{\cos }^{4}x-{\cos }^{2}x+1}$ is equal to

• A. $0$
• B. $1$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $1$

$\underset{x\to \infty }{lim}\frac{{\sin }^{4}x-{\sin }^{2}x+1}{{\cos }^{4}x-{\cos }^{2}x+1}$
$=\underset{x\to \infty }{lim}\frac{(1-{\cos }^{2}x{)}^2-(1-{\cos }^{2}x)+1}{{\cos }^{4}x-{\cos }^{2}x+1}$
$=\underset{x\to \infty }{lim}\frac{{\cos }^{4}x-{\cos }^{2}x+1}{{\cos }^{4}x-{\cos }^{2}x+1}=1$