Calculate the limit given $\displaystyle \lim_{x \to 0} \dfrac{\sin (x \sin \frac1x)} {x \sin \frac1x} $
Is it true that the answer is $1$, by  using the theorem of trigonometry limit? 
$\displaystyle \lim_{x \to 0} \dfrac{\sin x} {x}  = 1$
Do you have another way to strengthen the argument, L'Hospital like that? I have some doubt in this problem. 
 A: HINT
Note that
$$x \sin \frac1x\to 0$$
Indeed
$$-x\leq x \sin \frac1x \leq x$$
and for squeeze theorem the limit holds.
Thus consider $y= x \sin \frac1x\to 0$
$$\displaystyle \lim_{x \to 0} \dfrac{\sin (x \sin \frac1x)} {x \sin \frac1x}= \lim_{y \to 0} \dfrac{\sin y} {y} $$
A: It is true by limits of composition of continuous functions. 
If $\lim\limits_{x\to a} g(x)=b$ and $\lim\limits_{x\to b} f(x)=c$ then $\lim\limits_{x\to a}f(g(x))=c$
Real Analysis: Continuity of a Composition Function
Apply to $\begin{cases}g(x)=x\sin(\frac 1x) & a=0 & \lim\limits_{x\to 0}g(x)=b=0\\\\f(x)=\dfrac{\sin(x)}x & b=0 & \lim\limits_{x\to 0}f(x)=c=1\end{cases}$
Since $|\sin(\frac 1x)|\le 1$ you have $|g(x)|\le x\to 0$
A: Note that the function is undefined at some points in any deleted neighborhood of $0$ so if your textbook definition of limit requires that the function be defined in a deleted neighborhood of the point under consideration then the limit in question simply does not exist.
On the other hand if your textbook is mature enough to give definition of limit based on the pre-requisite that the point under consideration be a limit/accumulation/cluster point of the domain of the function then the limit in question exists and is equal to $1$ precisely via the argument you have provided.
