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$\lim\limits_{x\to 0} \frac{\sin(x\sin\frac{1}{x})}{x\sin\frac{1}{x}}$
Since $\lim x\sin\frac{1}{x} = 0$, and $\lim \frac{\sin x}{x} = 0$, final answer should be $1$.

Is the answer $1$ or it does not exist?

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    Use LaTeX formatting! Answer $1$ seems good for me. – 2017-02-15
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    You are right, and oscillations do no matter. (The argument of $\sin x/x$ can tend to zero the way it likes.) – 2017-02-15

1 Answers 1

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Hint:

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

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    What is $\bigodot$ mean? – 2017-02-15
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    @S.C.B.: quodlibet. – 2017-02-15
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    @quodlibet = Whatever you want, but of course not 0. But anything that converges to zero is ok. – 2017-02-15
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    Since $\frac{\sin(x\sin\frac{1}{x})}{x\sin\frac{1}{x}}$ is not defined in a neighborhood of $0$ (or even in a one-sided neighborhood of $0),$ it seems to me that the limit does not exist (as is ordinarily defined). – 2017-02-15