# Finf simple limit

1. Oct 9, 2009

### John O' Meara

Find $$\lim_{x \rightarrow 1} \frac{\sin(\pi x)}{x-1}$$. I have tried to find this limit by letting $$t= \pi x$$ t= x-1, etc. All I get is 1/0 or $$\pi/0$$ or etc., but not the answer the graph of the function suggests. I cannot find the substitution for x that will work. Is there some rule that I can use to find a suitable substitution for x or expression for t. I am doing this for my own interest. Can anyone point me along the correct line of reasoning that will allow me find the expression for t= ? . Thank you.

2. Oct 9, 2009

### JG89

Have you tried l'hopital's rule?

3. Oct 9, 2009

### John O' Meara

No, the book does not mention L'Hopital's rule until page 470, I am on page 135, so I don't think it means us to use that rule.

4. Oct 9, 2009

### JG89

$$\frac{sin(\pi x)}{x-1} = \frac{sin(\pi(x-1) + \pi)}{x-1} = \frac{sin(\pi(x-1))cos\pi + sin(\pi)cos(\pi(x-1))}{x-1} = \frac{-sin(\pi(x-1))}{x-1}$$.

Let $$t = x - 1$$. Then $$\frac{-sin(\pi(x-1))}{x-1} = \frac{-sin(\pi t)}{t} = \frac{-\pi sin(\pi t)}{\pi t}$$.

As x goes to 1, t = x - 1 goes to 0. You know the limit of sin(u)/u as u goes to 0 right? Now use that.

5. Oct 9, 2009

### John O' Meara

Thanks very much.