1. ## trig limits...help?

lim sin(2x)/tanx , x-->0

I'm not sure if I'm doing this right, I've never used one of these before...hopefully I'm posting this in the right spot.
Anyways, I can't seem to figure this question out...any help would be much appreciated!! Thanks

2. $\lim_{x\to 0}\frac{sin(2x)}{tan(x)}$

$\lim_{x\to 0}\frac{2sin(x)cos(x)}{\frac{sin(x)}{cos(x)}}$

$2\lim_{x\to 0}cos^{2}(x)=2$

3. Thanks for the quick reply...I'm just not quite sure how you got from one step to the next. I'm a little confused..?

4. Two identities are used:
• $\sin (2\theta) = 2\sin \theta \cos \theta$
• $\tan \theta = \frac{\sin \theta}{\cos \theta}$

So:
$\lim_{x \to 0} \frac{\sin 2x}{\tan x} = \lim_{x \to 0} \frac{\ 2\sin x \cos x \ }{\displaystyle \frac{\sin x}{\cos x}}$

Then do some algebra (dividing fractions is the same thing as multiplying the numerator by the reciprocal of the denominator):

$= \lim_{x \to 0 } 2\sin x \cos x \cdot \frac{\cos x}{\sin x}$

Things cancel and you should be good.

5. where does the cos(x) come from in the numerator?

6. Really not possible to explain it better...

7. so it's impossible to explain how that cos(x) just popped up into the numerator?

8. Not if you know that to divide by a fraction, you "invert and multiply"!

$\frac{a}{\frac{b}{c}}= a\left(\frac{c}{b}\right)= \frac{ac}{b}$
Do you understand that?

When James Bond said he couldn't explain it better, he had already said that:
"Then do some algebra (dividing fractions is the same thing as multiplying the numerator by the reciprocal of the denominator):"

9. Well, thank you. That's all I needed.