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The problem is:

$\lim_{x\to0}$ $\frac{sin(\frac{1}{x})}{sin(\frac{1}{sin(x)})}$

my intuition tells that the answer equal to $1$ by the limit equality : $\lim_{x\to0}$ $\frac{sinx}{x}=1$. But we can easily see that the limit of numerator and denominator both don't exist, and they are both between -1 and 1. I cannot seem to find a rigorous argument for this problem. If any one can help or give some hint would be very much appreciated!

superman
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  • I was thinking that $(1/x)/(1/sinx)= (sinx)/x$, and applying $sin$ to numerator and denominator would yield same answer – superman Nov 02 '16 at 19:51
  • If you switch to a sequence, say letting $x=1/n$ with $n=1,2,3,\ldots$, you might find $\sin(n)/\sin(1/\sin(1/n))\to1$ as $n\to\infty$.... – Barry Cipra Nov 02 '16 at 20:10

2 Answers2

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The limit doesn't exist because the function is not defined in any neighborhood of $0$. Indeed: $$x=\arcsin\left(\dfrac{1}{k\pi}\right), k\in\mathbb{Z}\Rightarrow\sin\left(\dfrac{1}{\sin(x)}\right)=0$$ and the sequence $\arcsin\left(\dfrac{1}{k\pi}\right)$ converges to $0$ as $k\rightarrow\infty$.

Olivier Moschetta
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  • Excuse me, but some authors require $x_0$ to be just a limit point of the domain of $f$ in the definition of the limit $lim_{x \to x_0} f(x)$. Perhaps what Barry Cipra commented might work. – Pythagoricus Nov 02 '16 at 20:32
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Let:

$$f(x) = \frac{\sin\left(\frac1x \right)}{\sin\left(\frac1{\sin x} \right)}$$

The limit doesn't exist because we have:

$$f\left(\frac2{(4n+1)\pi}\right) \not \to 0 \tag1$$

And

$$\lim_{n\to \infty} f\left(\frac1{2n\pi}\right) = 0$$

To prove $(1)$, note that the only way it could converge to $0$ is by having the denominator go to $\infty$, which is impossible because $|\sin| \le1$.