Calculating $ \lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) $

$$\lim_{x\to 0} \frac{ x\cdot \sin x}{|x|} = \lim_{x\to 0} \frac{ |x| \cdot \sin x}{x} = \lim_{x\to0} |x| \cdot \lim_{x\to0}\frac{\sin x}{x} = 0\cdot 1 = 0.$$


You can separate a limit into two only if both exist. Here as you point out, one of them does not, so this step is wrong.

You can say that $\operatorname{sgn}(x)$ is bounded, $\sin(x)$ goes to $0$, so their multiplication goes to $0$ as well.

Alternatively, calculate $x\to 0^+$ and $x\to 0^-$ and see they are equal.


Approach $0$ from the right:

$$\lim_{x \to 0^+} \frac{x\sin{x}}{|x|}=\lim_{x \to 0^+}\frac{x\sin{x}}{x}=\lim_{x \to 0^+}\sin x=0.$$

Then approach $0$ from the left:

$$\lim_{x \to 0^-} \frac{x\sin{x}}{|x|}=\lim_{x \to 0^-}\frac{x\sin{x}}{-x}=\lim_{x \to 0^-}-\sin x=0.$$

Since $\lim_{x \to 0^+}\frac{x\sin{x}}{|x|}=0=\lim_{x \to 0^-}\frac{x\sin{x}}{|x|}$, the limit is zero.

Tags:

Limits

Calculus

Solution Verification

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