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as $x$ approaches $1$.

$$\lim_{x\rightarrow 1} \frac{\sin |x-1|}{x-1} $$

I know that $x$ approaches $1$ from negative side and positive side but I don't know where to start it. I tried to cancel out the $x-1$ on the bottom but did not work.

Thanks.

Bruce Zheng
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jasminder88
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2 Answers2

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Hint: for $x<1, |x-1| = -(x-1)$ and for $x > 1, |x-1| = (x-1)$.

Also $\lim_{y \to 0} \frac{\sin y}{y} = 1$

What happens when one-sided limits are different?

Deepak
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Hints: Using $\lim_{x \to 0}\frac{\sin x}{x}=1$, we have

$$\lim_{ x\to 1^+}\frac{\sin|x-1|}{x-1}=\lim_{x\to 1^+}\frac{\sin(x-1)}{x-1}=1.$$

$$\lim_{ x\to 1^-}\frac{\sin|x-1|}{x-1}=\lim_{x\to 1^-}-\frac{\sin(x-1)}{x-1}=-1.$$

So its limit doesn't exist!

Paul
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  • In the first statement, that should be $x \to 0$, and perhaps it's better to use another variable like $y$ to avoid confusion (like in my answer). Also, that's the entire answer, not just "hints"! :) – Deepak Oct 15 '14 at 02:29