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$\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.

user
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Shane Dizzy Sukardy
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3 Answers3

6

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 $y= x \sin \frac1x\to 0$

$$\displaystyle \lim_{x \to 0} \dfrac{\sin (x \sin \frac1x)} {x \sin \frac1x}= \lim_{y \to 0} \dfrac{\sin y} {y} $$

user
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1

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 $|g(x)|\le x\to 0$

zwim
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  • Minor(?) quibble: What is the value of $f(g(x))$ at $x={1\over n\pi}$? – Barry Cipra Jan 06 '18 at 00:23
  • @BarryCipra I did not detailled the part about $f$ because it seemed known by OP that $f(0)=1$ and $f$ is continuous in $0$. So since $g(1/n\pi)=0$ it is expected to have $f(g(1/n\pi))=1$. We implicitly work with $\tilde g$ and $\tilde f$ the continuous extensions of $f$ and $g$. – zwim Jan 06 '18 at 00:25
1

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.

Paramanand Singh
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  • Thus we need to exclude previously the values for which $\sin \frac1x=0$ in order to guarantee the existence? – user Jan 06 '18 at 07:45
  • @gimusi : most definitions in analysis textbooks disregard such exceptional points and the only pre-requisite for a limit $\lim_{x\to a} f(x) $ is that $a$ be an accumulation point of domain of $f$. So in this question it is more a matter of convention that the limit exists or not. More experienced users follow the conventions based on accumulation point. – Paramanand Singh Jan 06 '18 at 08:15