Pedantic on Mathematical Software [Trap Limit]
Wolfram Alpha result for
\[\lim_{x\rightarrow 0}\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}\]
is "No result found in terms of standard mathematical functions" while Maple clearly says the limit is "Undefined". Notice that Wolfram Alpha will say "The Limit Does Not Exist" for $\lim_{x\rightarrow 0} \sin \left(\frac{1}{x}\right)$.
Indeed Maple is right, the limit does not exist. It is important to notice that the following solution is wrong
\[\lim_{x\rightarrow 0} x\sin \frac{1}{x}=0 \qquad \text{and} \qquad \lim_{x\rightarrow 0} \frac{\sin{x}}{x}=1\] Then \[\lim_{x\rightarrow 0}\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}=\lim_{x\sin \frac{1}{x} \rightarrow 0} \frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}=\lim_{y\rightarrow 0} \frac{\sin y}{y}=1\]
First, we cannot use the composition law of limit here, the law says that:
[Substitution Rule] Suppose that as $x\rightarrow a$, $g(x)$ converges properly to $A$ and that \[\lim_{y\rightarrow A} f(y)=L.\]
Then
\[\lim_{x\rightarrow a} f(g(x))=L\]
\[\lim_{x\rightarrow 0} x\sin \frac{1}{x}=0 \qquad \text{and} \qquad \lim_{x\rightarrow 0} \frac{\sin{x}}{x}=1\] Then \[\lim_{x\rightarrow 0}\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}=\lim_{x\sin \frac{1}{x} \rightarrow 0} \frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}=\lim_{y\rightarrow 0} \frac{\sin y}{y}=1\]
First, we cannot use the composition law of limit here, the law says that:
[Substitution Rule] Suppose that as $x\rightarrow a$, $g(x)$ converges properly to $A$ and that \[\lim_{y\rightarrow A} f(y)=L.\]
Then
\[\lim_{x\rightarrow a} f(g(x))=L\]
Though, $\lim_{x\rightarrow 0} x \sin\frac{1}{x}=0$, the convergent is not proper, thus the above rule can't be applied.
To see why the limit does not exist, We look back at the definition of the limit, we say that such a limit exists if we can find a real number $L$, such that:
\[\forall \epsilon > 0 \exists \delta > 0 , \forall x, \, \, \, 0< |x-c| < \delta \Rightarrow \left|f(x)-L\right| < \epsilon \]
The problem is: for every $L \in \mathbb{R}$, given an $\epsilon > 0$, no matter how small we choose a neighbourhood of $0$ (by choosing a sufficiently small $\delta$), the function $x\sin \frac{1}{x}$ can be equal to zero for infinitely many values of $x$ on that neighbourhood, thus the function
\[\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}\]
is undefined (infinitely many times) on that neighbourhood.
In other words, it is impossible for us to choose a neighbourhood around $0$ such that the function
\[\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}\]
is defined on all points on that neighbourhood, hence no such $\delta$ can be chosen on the definition of limit, thus the limit does not exist.
More formally, for any $\delta > 0$, by Archimedean Principle there exists a positive integer $N$ such that $N>\frac{1}{2\pi \delta}$, thus we have $x_k=\frac{1}{2k\pi} \in (0,\delta)$ for every $k\geq N$, and on these points we have $x_k \sin \frac{1}{x_k} = 0$.
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