• Mar 27th 2009, 04:06 PM
Kaitosan
How is it possible that lim(x to infinity) sin(x/2)/x = 1/2? I thought it was zero.....? After all, the maximum value of the numerator is 1 in absolute value while the denominator approaches infinity, thus making the expression approaching zero.

So what's up guys? What am I missing?
• Mar 27th 2009, 04:13 PM
skeeter
Quote:

Originally Posted by Kaitosan
How is it possible that lim(x to infinity) sin(x/2)/x = 1/2? I thought it was zero.....? After all, the maximum value of the numerator is 1 in absolute value while the denominator approaches infinity, thus making the expression approaching zero.

So what's up guys? What am I missing?

$\displaystyle \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{x} = \frac{1}{2}$

not $\displaystyle x \to \infty$

must be a misprint.
• Mar 27th 2009, 04:48 PM
matheagle
$\displaystyle \lim_{x\to\infty}{\sin x\over x}=0$ while $\displaystyle \lim_{x\to 0}{\sin x\over x}=1$

giving you $\displaystyle \lim_{x\to 0}{\sin(x/2)\over x}=1/2$

SO which one do you want?
• Mar 27th 2009, 09:52 PM
Kaitosan
Guys! I didn't misprint.

It's lim(x to infinity) sin(x/2)/x. The book must be wrong then.
• Mar 27th 2009, 09:59 PM
matheagle
Then $\displaystyle {-1\over x} \le {\sin (x/2)\over x} \le {1\over x}$ and since $\displaystyle {1\over x}\to 0$ as $\displaystyle x\to \infty$ the limit is zero.
• Mar 28th 2009, 09:37 AM
stapel
If the limit is being taken as x goes to infinity, then you are correct.

I suspect that there is a typo in the book, and that the limit was supposed to have been "as x tends toward zero". If so, then the book's answer would have been correct. (Wink)