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How do I take limits of the form $$\lim_{x \rightarrow 1} \left((x - 1)\sin\left(\frac{x}{x-1}\right)\right)$$?

Using boundedness of $\sin$ seems wrong.

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Use boundedness of sin. –  akkkk Nov 8 '12 at 15:01
    
If you have a reason something "seems wrong" then that usually points you to a disproof. However if you don't have a reason then you might want to check the details, because it might actually work! –  rschwieb Nov 8 '12 at 15:07
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2 Answers

up vote 1 down vote accepted

Clearly $$- (x - 1) \leq (x - 1)\sin \left( {\frac{x}{{x - 1}}} \right) \leq (x - 1)$$ and $\pm(x - 1) \to 0$ as $x \to 1$, so by the squeeze theorem we know that $$(x - 1)\sin \left( {\frac{x}{{x - 1}}} \right) \to 0.$$

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Why does using boundedness seem wrong?

We have for all $x \in \mathbb R$ that \[ 0 \le \left|(x-1)\sin\frac x{x-1}\right| \le \left|x-1\right| \] As $\lim_{x\to 1} 0 = \lim_{x\to 1} \left|x-1\right| = 0$, we have \[ \lim_{x\to 1} (x-1)\sin\frac x{x-1} = 0 \] by the sandwich theorem.

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