Q. 76
Question
Prove that the converse of Theorem \(7.9\) is not true by finding a continuous function \(a:\left [ 1,\infty \right )\rightarrow R\) such that \(\lim_{x\rightarrow \infty }a\left \( x \right \)\) does not exist but \(\left \{ a\left \( x \right \) \right \}\) converges.
Step-by-Step Solution
VerifiedAn example of a continuous function \(a:\left [ 1,\infty \right )\rightarrow R\) such that \(\lim_{x\rightarrow \infty }a\left \( x \right \)\) does not exist but \(\left \{ a\left \( x \right \) \right \}\) converges is \(a\left ( x \right )=\sin \left ( \pi x \right )\).
The objective is to find an example of a continuous function \(a:\left [ 1,\infty \right )\rightarrow R\) such that \(\lim_{x\rightarrow \infty }a\left \( x \right \)\) does not exist but \(\left \{ a\left \( x \right \) \right \}\) converges.
Consider the function \(a\left ( x \right )=\sin \left ( \pi x \right )\).
The function is continuous on \(\left [ 1,\infty \right )\) and \(\lim_{x\rightarrow \infty }a\left \( x \right \)\) does not exist.
CASE \(1\): \(x \epsilon \mathbb{Z}\) then
\(\begin{align*}
a\left \{ x \right \} &= \sin\left ( \pi x \right ) \\
a\left \{ x \right \} &= 0
\end{align*}\)
Then, (\lim_{x\rightarrow \infty }a\left \( x \right \)=0\).
Case \(2\): If x=\frac{4n+1}{2}, n\epsilon \mathbb{N} then
\(\begin{align*}
a\left \{ x \right \} &= \sin\left ( \pi x \right ) \\
a\left \{ x \right \} &= \sin\left ( \frac{4n+1}{2}\pi \right ) \\
a\left \{ x \right \} &= 1
\end{align*}\)
and
\(\begin{align*}
\lim_{x\rightarrow \infty }a\left \{ x \right \} &= \lim_{x\rightarrow \infty } \sin\left ( \pi x \right ) \\
&= \lim_{x\rightarrow \infty } \sin\left ( \frac{4n+1}{2}\pi \right ) \\
&= 1
\end{align*}\)
So, it is clear that limit does not exist.
Now,
\(\begin{align*}
a_{k}&=a(k) \\
& =\sin\left (k\pi \right )
\end{align*}\)
As, \(k\epsilon \mathbb{N}\) so \(\sin\left (k\pi \right )=0\).
As each term of the series is \(0\), so \left \(\{ a\left ( x \right ) \right \}\) converges to \(0\).