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$$\lim_{x \to 0} x \cdot \sin{\frac{1}{x}}\cos{\frac{1}{x}}$$

I don't solve this kind of limits, I can't try anything because it seems difficult to me.

Kamil Jarosz
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user315918
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  • HINT: since $\sin, \cos$ are bounded $$-|x| \le x \sin (1/x) \cos (1/x) \le |x|$$ – Crostul Mar 08 '16 at 15:05
  • @user315918 don't forget to accept one of the answers clicking the 'V' in the left side of the answer, and also to vote – 3SAT Mar 08 '16 at 19:27

3 Answers3

1

The answer is zero because $\sin \theta$ and $\cos \theta$ are bounded

$$-1\le\sin(1/x)\le 1\\ -1\le\cos(1/x)\le 1$$

$$\lim\limits_{x\to 0}\color{green}x=0$$

You have "zero$\times$ bounded"$=0$

$$\Longrightarrow\lim_{x \to 0} \color{green}x \cdot \sin{\frac{1}{x}}\cos{\frac{1}{x}}=\color{red}0$$

3SAT
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1

HINT:

As for real $x,$ $$-1\le\sin\dfrac2x=2\sin\dfrac1x\cos\dfrac1x\le1$$

lab bhattacharjee
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0

Let $\epsilon>0$ be given. Now consider $ |x\sin \frac{1}{x}\cos\frac{1}{x}|\leq|x|<\delta=\epsilon$, since sin and cos functions are bounded. So, the required limit is equal to 0.

Singh
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