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I'm trying to find the following limit:

$$ \lim_ {x \to 1} \frac{\sin{\pi x}}{1 - x^2} $$

I can't figure it out how to reach the fundamental trigonometric limit. Everything i see is that the denominator is a difference between squares, and then can be factorated

$$ \lim_ {x \to 1} \frac{\sin{\pi x}}{(1 - x)(1 + x)} $$

I'd like to know how can i simplify this expression to eliminate the indetermination.

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    $\begingroup$ Hint: $\sin\theta= -\sin( \theta-\pi)=\sin(\pi-\theta)$. $\endgroup$ Dec 19, 2012 at 0:18

2 Answers 2

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Let $y = 1-x$. Then we get that \begin{align} \lim_{x \to 1}\dfrac{\sin(\pi x)}{(1-x)(1+x)} & =\lim_{y \to 0}\dfrac{\sin(\pi (1-y))}{y(2-y)}\\ & =\lim_{y \to 0}\dfrac{\sin(\pi y)}{y} \dfrac1{(2-y)}\\ & =\lim_{y \to 0}\dfrac{\sin(\pi y)}{y} \cdot \lim_{y \to 0} \dfrac1{(2-y)}\\ & = \pi \cdot \dfrac12 & \left(\text{Recall that}\lim_{y \to 0}\dfrac{\sin(a y)}{y} = a \right)\\ & = \dfrac{\pi}{2} \end{align}

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You can also appeal to L'Hopital's Rule:

$$ \lim_{x \to 1}\frac{\sin(\pi x)}{(1-x^2)} =\lim_{x \to 1}\frac{\pi \cos(\pi x)}{-2 x} \\ = \frac{\pi}{2}$$

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