# One more limit with trig

1. Feb 4, 2004

### burge

Stuck here too:

lim as x -> 0 of [cotPxsinx]/2secx
*P = pi

Last edited: Feb 4, 2004
2. Feb 4, 2004

### himanshu121

This is equivalent to

$$\lim_{x\rightarrow 0} = \frac{sinx}{2sin\pi x}$$

Last edited: Feb 4, 2004
3. Feb 4, 2004

### burge

How did you get to that?

4. Feb 4, 2004

### himanshu121

cotpix=cospix/sinpix.
so therefore u have
$$\lim_{x\rightarrow 0} = \frac{sinx}{2sin\pi x}$$
+ Using these
cosx->1 as x->0

5. Feb 5, 2004

### HallsofIvy

Staff Emeritus
In more detail:

$$cot(\pi x)= \frac{cos(\pi x)}{sin(\pi x)}$$ and
sec(x)= $$\frac{1}{cos(x)}$$

so
$$\frac{cot(\pi x)sin(x)}{sec(x)}= \frac{cos(\pi x)sin(x)}{sin(\pi x)cos(x)}$$
$$= \frac{cos(\pi x)}{cos(x)}\frac{sin(x)}{sin(\pi x)}$$

Since both $$cos(\pi x)$$ and cos(x) have limit 1 as x-> 0, the limit of $$\frac{cos(\pi x)}{cos(x)}= 1$$ and we are left with
$$limit_{x->0}\frac{sin(x)}{sin(\pi x)}$$.

You could do that by L'hopital's rule or by considering the first few terms of the Taylor's series for sine.