one more limit with trig

[burge]

Stuck here too:

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

*Thanks for your help

[himanshu121]

This is equivalent to

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

[burge]

How did you get to that?

[himanshu121]

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

[HallsofIvy]

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.