the function f is given by f(x)=sin((x + 1)/x²). which of the following statements are true? i. the graph of…

Explanation:

Step1: Analyze horizontal asymptotes

Find $\lim_{x\rightarrow\pm\infty}\sin\left(\frac{x + 1}{x^{2}}\right)$. As $x\rightarrow\pm\infty$, we consider the behavior of $\frac{x + 1}{x^{2}}=\frac{1}{x}+\frac{1}{x^{2}}$. Then $\lim_{x\rightarrow\pm\infty}\left(\frac{1}{x}+\frac{1}{x^{2}}\right)=0$. Since $\sin$ is a continuous - function, $\lim_{x\rightarrow\pm\infty}\sin\left(\frac{x + 1}{x^{2}}\right)=\sin(0) = 0$. So, $y = 0$ is a horizontal asymptote.

Step2: Check if $y = 1$ is a horizontal asymptote

We have already shown that $\lim_{x\rightarrow\pm\infty}\sin\left(\frac{x + 1}{x^{2}}\right)=0$. Since $\sin\left(\frac{x + 1}{x^{2}}\right)$ approaches $0$ as $x\rightarrow\pm\infty$, $y = 1$ is not a horizontal asymptote.

Step3: Analyze vertical asymptotes

The function $f(x)=\sin\left(\frac{x + 1}{x^{2}}\right)$ is undefined when $x^{2}=0$, i.e., $x = 0$. As $x\rightarrow0$, $\frac{x + 1}{x^{2}}\rightarrow+\infty$. The sine - function oscillates as its argument approaches infinity. So, $x = 0$ is a vertical asymptote.

Answer:

D. I and III only