# Solving this trig limit without L'hopital: $\lim_{x\to 1} \frac{\sin(\pi x)}{x-1}$?

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?

• Let $x=y+1$ first – Claude Leibovici May 4 '17 at 11:12
• 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. – Umberto P. May 4 '17 at 11:13
• You can do it by simple calculations: Evaluate the expression on the left of $1$ ($0.999$), and on the right of $1$ ($1.001$). – Toby Mak May 4 '17 at 11:14
• @TobyMak that method can only give you an approximate answer. – GFauxPas May 4 '17 at 11:49
• I marked the answer of @TheDeadLegend as answer, but all of the answers helped me understand the point, – twkmz May 4 '17 at 12:06

$$\lim_{h\rightarrow0}\frac{\sin(\pi(1-h))}{1-1-h}$$ $$\lim_{h\rightarrow0}\frac{\sin(\pi h)\pi}{-\pi h}\tag{Multiply and divide by pi}$$ $$-\pi$$

• thanks, this really helped me understand the exercise. just for clarification $$sin(x) = sin(\pi - x)$$ right? – twkmz May 4 '17 at 12:04
• Yes. Exactly @twkmz – The Dead Legend May 4 '17 at 12:05

\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}

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.)

With equivalents:

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

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