1. ## Limits

Hi!

I need some help to get me started on these $\displaystyle \lim_{x\rightarrow 0}$ expressions.

I just want to learn the mindset behind solving these, so I can try to solve a few on my own.

$\displaystyle \lim_{x\rightarrow 0} \ \frac{sin 3x}{3x}$

I want to decide the limit when $\displaystyle x\rightarrow \ 0$

Thank you

2. Originally Posted by Twig
Hi!

I need some help to get me started on these $\displaystyle \lim_{x\rightarrow 0}$ expressions.

I just want to learn the mindset behind solving these, so I can try to solve a few on my own.

$\displaystyle \lim_{x\rightarrow 0} \ \frac{sin 3x}{3x}$

I want to decide the limit when $\displaystyle x\rightarrow \ 0$

Thank you
Just remember that $\displaystyle \lim_{u\to{0}}\frac{\sin(u)}{u}=1$

That should pretty much lead you to the answer you are looking for.

--Chris

3. ## hi

Hi!

Uhm.. letīs see here.

$\displaystyle \lim_{x\rightarrow 0} \ \frac{sin 3x}{3x}$

If I let $\displaystyle 3x = v$

For every positive number $\displaystyle \epsilon$ there is a positive number
$\displaystyle \delta$ so that $\displaystyle 0 < |\alpha - 0| < \delta$ so that $\displaystyle |sin (\alpha) - 1| < \epsilon$

Meaning in this case that for every angle v, $\displaystyle sin (v) = \ \mbox{some value}$
,there is always an angle closer to 0 that makes $\displaystyle sin (\alpha) = \ \mbox{lower value}$

Is this correct?

I donīt really get how to proceed...

$\displaystyle sin (v) = 1 - \epsilon$
then I know from the definitions above that $\displaystyle 0 < sin (\alpha) < 1 - \epsilon$ , since $\displaystyle \alpha \ \mbox{is an angle closer to 0, thus making the sine of the angle closer to 0}$

From here I donīt really know what to do...

Thanks