# trig limits...help?

• February 20th 2009, 06:06 PM
edge
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
• February 20th 2009, 06:20 PM
galactus
$\lim_{x\to 0}\frac{sin(2x)}{tan(x)}$

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

$2\lim_{x\to 0}cos^{2}(x)=2$
• February 20th 2009, 06:40 PM
edge
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..?
• February 20th 2009, 06:48 PM
o_O
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.
• February 21st 2009, 06:40 AM
edge
where does the cos(x) come from in the numerator?
• February 21st 2009, 07:57 AM
james_bond
Really not possible to explain it better...
• February 21st 2009, 08:12 AM
edge
so it's impossible to explain how that cos(x) just popped up into the numerator?
• February 21st 2009, 08:24 AM
HallsofIvy
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):"
• February 21st 2009, 09:09 AM
edge
Well, thank you. That's all I needed.