Estimate the horizontal asymptote. $y = \frac { 1 - \cos ^ { 2 } \left( \frac { 2 } { x } \right) } { \sin \left( \frac { 9 } { x } \right) }$
A) $y = 9$
B) $y = 4$
C) $y = 2$
D) $$y = 0$$
E) $y = - 9$

Question

Quizplus · Accepted Answer

As $x \to \infty$, $\cos^2 \left( \frac{2}{x} \right) \to 1$ and $\sin \left( \frac{9}{x} \right) \to 0$. So, $y \to \frac{1-1}{0} = \frac{0}{0}$. Applying L'Hopital's Rule, we get$$\lim_{x\to\infty} \frac{1-\cos^2 \left( \frac{2}{x} \right)}{\sin \left( \frac{9}{x} \right)} = \lim_{x\to\infty} \frac{\frac{d}{dx} \left[ 1-\cos^2 \left( \frac{2}{x} \right) \right]}{\frac{d}{dx} \left[ \sin \left( \frac{9}{x} \right) \right]} = \lim_{x\to\infty} \frac{2\sin \left( \frac{2}{x} \right) \left( -\frac{2}{x^2} \right)}{9\cos \left( \frac{9}{x} \right) \left( -\frac{9}{x^2} \right)} = \lim_{x\to\infty} \frac{4\sin \left( \frac{2}{x} \right)}{81\cos \left( \frac{9}{x} \right)} = 0$$

Estimate the Horizontal Asymptote y=1−cos⁡2(2x)sin⁡(9x)y = \frac { 1 - \cos ^ { 2 } \left( \frac { 2 } { x } \right) } { \sin \left( \frac { 9 } { x } \right) }y=sin(x9​)1−cos2(x2​)​

Estimate the Horizontal Asymptote $y = \frac { 1 - \cos ^ { 2 } \left( \frac { 2 } { x } \right) } { \sin \left( \frac { 9 } { x } \right) }$