anonymous
  • anonymous
limit x->0, tan2x/x find the limit, help!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
l'Hopital's rule top and bottom. (2sec(x)^2)/1, limx->0 of 2sec(x)^2
anonymous
  • anonymous
we havent learned that rule, yet, so u wouldnt have tan=cos/sin?
anonymous
  • anonymous
sin/cos*

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TuringTest
  • TuringTest
is it\[\tan(2x)\]or\[\tan^2(x)\]?
anonymous
  • anonymous
tan2x not ^2
anonymous
  • anonymous
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?
anonymous
  • anonymous
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?
anonymous
  • anonymous
http://mathworld.wolfram.com/LHospitalsRule.html Apparently I spelled it wrong, but w/e. The proof is lovely.
anonymous
  • anonymous
you could try \[\lim_{x\to 0}\frac{\sin(2x)}{x\cos(2x)}\]
anonymous
  • anonymous
thats as far as i got what do i after that satel.?
anonymous
  • anonymous
break into to parts \[\lim_{x\to 0}\frac{1}{\cos(2x)}= 1\] and \[\lim_{x\to 0}\frac{\sin(2x)}{x}=2\]
TuringTest
  • TuringTest
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...
anonymous
  • anonymous
yeah but you are quicker because i did not explain my second limit
anonymous
  • anonymous
Man, algebra. I forget that if I can't solve a problem, it's probably just me doing algebra badly.
Zarkon
  • Zarkon
I would write \[\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}\]
TuringTest
  • TuringTest
dang...lol
anonymous
  • anonymous
or replace x by x/2
anonymous
  • anonymous
turning test, so 2cos^2x-1 cancels the top, n ur left wit 2?
TuringTest
  • TuringTest
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
anonymous
  • anonymous
the answer is 2 tho
TuringTest
  • TuringTest
that's what you get in total\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{\cos(2x)}=2\cdot1\cdot1=2\]
TuringTest
  • TuringTest
I didn't write the limits in ...lazy I know
anonymous
  • anonymous
got it!!

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