$$\lim_{x \rightarrow \infty}x \times\lim_{x \rightarrow \infty} \sin \frac{1}{x}=\infty \times0=Undefined$$ Is this the correct way to convey that the limit does not exist? Or is there a mathematical way to show that $$\lim_{x \rightarrow \infty} \sin \frac{1}{x} = 0$$ Other than just knowing that 1 divide by an infinitely large number approaches $0$. |
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You can only use the fact that $$\lim_{x \to \infty}f(x)g(x)=\lim_{x \to \infty} f(x) \lim_{x \to \infty}g(x)$$ When it is given both limits exist. So your method of saying this is undefined is incorrect. So the way to evaluate this limit is by setting $\frac{1}{x}=t$, which yields that $$\lim_{x \to \infty} x\sin \frac{1}{x} =\lim_{t \to 0^{+}} \frac{\sin t}{t}=1$$ |
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This doesn't actually demonstrate that the limit doesn't exist; it demonstrates that the simplest possible 'rule' for establishing the limit won't work. The key thing here is to realize that this is the same as $$ \lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}=\lim_{w\to0^+}\frac{\sin w}{w}=\lim_{w\to0^+}\frac{\sin w -\sin 0}{w-0}, $$ which is a special form you may recognize. |
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