Question

limit x->0, tan2x/x find the limit, help!!

Answer 1

l'Hopital's rule top and bottom. (2sec(x)^2)/1, limx->0 of 2sec(x)^2

Answer 2

we havent learned that rule, yet, so u wouldnt have tan=cos/sin?

Answer 3

sin/cos*

Answer 4

is it\[\tan(2x)\]or\[\tan^2(x)\]?

Answer 5

tan2x not ^2

Answer 6

I wonder, I don't think it's possible to solve tan(2x)/x without l'Hopital's. Unless I'm derping. Maybe some weird half angle identity?

Answer 7

my calc class just started last week n my professor has mentioned that rule but has yet to teach it, so wa does that rule state?

Answer 8

http://mathworld.wolfram.com/LHospitalsRule.html

Apparently I spelled it wrong, but w/e. The proof is lovely.

Answer 9

you could try
\[\lim_{x\to 0}\frac{\sin(2x)}{x\cos(2x)}\]

Answer 10

thats as far as i got what do i after that satel.?

Answer 11

break into to parts
\[\lim_{x\to 0}\frac{1}{\cos(2x)}= 1\] and
\[\lim_{x\to 0}\frac{\sin(2x)}{x}=2\]

Answer 12

How about\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{2\cos^2x-1}=2\]sat beat me as usual...

Answer 13

yeah but you are quicker because i did not explain my second limit

Answer 14

Man, algebra.

I forget that if I can't solve a problem, it's probably just me doing algebra badly.

Answer 15

I would write \[\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}\]

Answer 16

dang...lol

Answer 17

or replace x by x/2

Answer 18

turning test, so 2cos^2x-1 cancels the top, n ur left wit 2?

Answer 19

I didn't even need to do that identity, don't know why I did
but you can just plug in x=0 to that second part and get 1

Answer 20

the answer is 2 tho

Answer 21

that's what you get in total\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{\cos(2x)}=2\cdot1\cdot1=2\]

Answer 22

I didn't write the limits in
...lazy I know

Answer 23

got it!!