Re: Question about limits

Quote:

Originally Posted by

**NecroWinter** the question involves lim x->0 (x cot x)

Sorry if the notation isnt so good but its the limit of x as it approaches 0

the book goes on to prove that the answer is 1 by doing the following steps that confuse me:

lim(x cot x) = lim(x (cosx/sin x)) =** lim ((x/sin x)cos x)**

=(lim(x/sin x))(lim cos x)

=((lim cos x)/(lim (sin x/x))) = 1/1 = 1

the bold part highlights my confusion, I dont understand what happened there, why the denominator went to x, and was removed by cos x

its very possible its just a simple mistake on my part, but for some reason im just lost.

thanks, and sorry about the notation, I dont know how to use the forum that well

$\displaystyle x\cot{x} = x\cdot \frac{\cos{x}}{\sin{x}} = \frac{x}{\sin{x}} \cdot \cos{x} = \cos{x} \cdot \frac{x}{\sin{x}} = \cos{x} \div \frac{\sin{x}}{x}$

Re: Question about limits

$\displaystyle x\frac{\cos(x)}{\sin(x)}=\frac{x\cos(x)}{\sin(x)}= \frac{x}{\sin(x)}\cos(x)$