# Thread: Limit help please!

1. ## Limit help please!

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?

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

not $x \to \infty$

must be a misprint.

3. $\lim_{x\to\infty}{\sin x\over x}=0$ while $\lim_{x\to 0}{\sin x\over x}=1$

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

SO which one do you want?

4. Guys! I didn't misprint.

It's lim(x to infinity) sin(x/2)/x. The book must be wrong then.

5. Then ${-1\over x} \le {\sin (x/2)\over x} \le {1\over x}$ and since ${1\over x}\to 0$ as $x\to \infty$ the limit is zero.

6. 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.

Ask your instructor for correction or clarification.