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 $1$

For all $x$,
${sin}^{4}x={\left({sin}^{2}x\right)}^2$
$={(1-{cos}^{2}x)}^2$
$=1-2{cos}^{2}x+{cos}^{4}x$
Thus numerator $=1-2{cos}^{2}x+{cos}^{4}x+{cos}^{2}x=1-{cos}^{2}x+{cos}^{4}x$
Hence, for all x, numerator $=$ denominator.
Thus the limit $=1$ for all $x$