Answer
(a) No; $f(n)$ oscillates between $\pm$1 and $0$ .
(b) $-1,+1,-1,+1,-1$
(c) No, it oscillates between $+1$ and $-1$ .
Work Step by Step
(a) To determine if $\displaystyle\lim _{x \rightarrow+\infty} f(x)$ exists, we need to consider the behavior of $f(x)$ as $x$ approaches infinity. Note that $\cos \left(\frac{\pi}{2} x\right)$ has a period of $4$, which means that $f(x)$ will repeat itself every $4$ units along the $\mathrm{x}$-axis. As $x$ becomes very large positive values, the cosine function will oscillate between $-1$ and $1$ with increasing frequency. Therefore, the limit does not exist.
(b) To evaluate $a_1, a_2, a_3, a_4$, and $a_5$, we substitute the given values of $n$ into the expression for $a_n=f(2 n)$ :
$\begin{align*}
a_1 &= f(2(1)) = f(2) = \cos(\pi) = -1 \\
a_2 &= f(2(2)) = f(4) = \cos(2\pi) = 1 \\
a_3 &= f(2(3)) = f(6) = \cos(3\pi) = -1 \\
a_4 &= f(2(4)) = f(8) = \cos(4\pi) = 1 \\
a_5 &= f(2(5)) = f(10) = \cos(5\pi) = -1 \\
\end{align*}$
(c) To determine if $\left\{a_n\right\}$ converges, we need to check if it is a Cauchy sequence. Let's consider the difference between two arbitrary terms of the sequence:
$\begin{align*}
|a_n - a_m| &= |f(2n) - f(2m)| \\
&= |\cos(\frac{\pi}{2}(2n)) - \cos(\frac{\pi}{2}(2m))| \\
&= |\cos(\pi n) - \cos(\pi m)| \\
&= 2|\sin(\frac{\pi(n+m)}{2})\sin(\frac{\pi(n-m)}{2})|
\end{align*}$
Since $\left|\sin \left(\frac{\pi(n+m)}{2}\right)\right|$ and $\left|\sin \left(\frac{\pi(n-m)}{2}\right)\right|$ are both bounded by 1 , we have:
Since the difference between any two terms is bounded above by a constant, we can conclude that $\left\{a_n\right\}$ is a Cauchy sequence. However, since $a_n$ oscillates between $- 1$ and $1$ depending on whether $n$ is even or odd, the sequence does not converge.
