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I know similar questions have been asked but I could not find any answer that applies to my case, so I'll ask anyway. I'm having trouble getting rid of an indeterminate without using derivatives

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

One idea I had was to multiply both sides of the fractions by $$\pi x$$ like this $$\lim_{x\to 1} \frac{\sin(\pi x)\pi x}{\pi x^2-\pi x}$$ to cancel out the sine but then I realized that I could not do that because x tends to 1, and not to 0. What can I do?

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  • $\begingroup$ Let $x=y+1$ first $\endgroup$ – Claude Leibovici May 4 '17 at 11:12
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    $\begingroup$ Can you use calculus? Your limit is the same as $$\lim_{x \to 1} \frac{\sin \pi x - \sin \pi}{x-1}$$ which should hopefully look familiar. $\endgroup$ – Umberto P. May 4 '17 at 11:13
  • $\begingroup$ You can do it by simple calculations: Evaluate the expression on the left of $1$ ($0.999$), and on the right of $1$ ($1.001$). $\endgroup$ – Toby Mak May 4 '17 at 11:14
  • $\begingroup$ @TobyMak that method can only give you an approximate answer. $\endgroup$ – GFauxPas May 4 '17 at 11:49
  • $\begingroup$ I marked the answer of @TheDeadLegend as answer, but all of the answers helped me understand the point, $\endgroup$ – twkmz May 4 '17 at 12:06
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$$\lim_{h\rightarrow0}\frac{\sin(\pi(1-h))}{1-1-h}$$ $$\lim_{h\rightarrow0}\frac{\sin(\pi h)\pi}{-\pi h}\tag{Multiply and divide by pi}$$ $$-\pi$$

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  • $\begingroup$ thanks, this really helped me understand the exercise. just for clarification $$sin(x) = sin(\pi - x)$$ right? $\endgroup$ – twkmz May 4 '17 at 12:04
  • $\begingroup$ Yes. Exactly @twkmz $\endgroup$ – The Dead Legend May 4 '17 at 12:05
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\begin{eqnarray} \lim_{x \rightarrow 1}\frac{\sin(\pi x)}{1-x} &=&\lim_{h \rightarrow 0}\frac{\sin(\pi (1-h))}{h} \\ &=&\lim_{h \rightarrow 0}\frac{\sin(-\pi h)}{h} \\ &=&\lim_{h \rightarrow 0}\frac{-\sin(\pi h)\pi}{\pi h} \tag{common limit} \\ &=&-\pi. \end{eqnarray}

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Substitute $\pi x=t+\pi$, so $x-1=\frac{t}{\pi}$; for $x=1$ we have $t=0$, so the limit becomes $$ \lim_{x\to1}\frac{\sin(\pi x)}{x-1}= \lim_{t\to0}\frac{\sin(t+\pi)}{t/\pi}= \lim_{t\to0}-\pi\frac{\sin t}{t} $$


On the other hand the limit is the derivative at $1$ of the function $f(x)=\sin(\pi x)$, and $$ f'(x)=\pi\cos(\pi x) $$ so $$ f'(1)=-\pi $$

(Note: this is not l’Hôpital.)

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With equivalents:

Set $x=1+h$. We know $\sin u\sim_0u$, so $$\frac{\sin\pi x}{x-1}=\frac{\sin(\pi+\pi h)}{h}=-\frac{\sin\pi h}h\sim_0-\frac{\pi h}h=-\pi.$$

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

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