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Finding Limit of Trig Func 
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#1
Oct911, 12:32 AM

P: 1,306

1. The problem statement, all variables and given/known data
[itex]\lim_{x\to\0} \frac{2tan^2x}{x}[/itex] 3. The attempt at a solution [itex]\lim_{x\to\0} \frac{2tan^2x}{x} \\ = \frac{2 tanx tanx}{x} \\ = \cfrac{2 \cfrac {sin}{cos} \cfrac {sin}{cos}[/itex] ? Edit: Oh boy none of my latex is working. :( 1. The problem statement, all variables and given/known data lim x> 0 (2tan^2x)/x 3. The attempt at a solution lim x> 0 (2tan^2x)/x = [ 2 tanx (tan x) ] / x = [ 2 (sin / cos) (sin/cos) ] / x = [ 2 (sin^2x/cos^2x) ] / x = [ 2sin^2x / cos^2x ] \ x Help please. Knowing me the answer is probably pretty simple. =D 


#2
Oct911, 01:02 AM

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Do you know the limit x>0 (sin(x)/x)?
ehild 


#3
Oct911, 01:12 AM

P: 1,306

lim x> 0 (2tan^2x)/x = [ 2 tanx (tan x) ] / x = [ 2 (sin / cos) (sin/cos) ] / x From here on out, I decided to multiply the numerator and denominator by cos/sin = [ [ 2 (sin / cos) (sin/cos) ] / x ] cos/sin = [2 (cos/sin) ] / (cos/sin) = 2 Can anyone confirm this? Its an even number in the book so I can't tell if I get it right or wrong, and cramster.com doesn't supply this problem. Also any ideas why my latex code didn't work? 


#4
Oct911, 01:39 AM

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Finding Limit of Trig Func
Your idea does not help and the result is wrong.
ehild 


#5
Oct911, 06:45 AM

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[tex]\lim_{x \to 0} \frac{tan^2(x)}{x}=\lim_{x \to 0} \frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}[/tex]
ehild 


#6
Oct911, 07:32 AM

P: 1,195

Do you know l'Hopital's rule?



#7
Oct911, 02:17 PM

P: 1,306




#8
Oct911, 02:28 PM

P: 1,306

[tex] \lim_{x \to 0} \frac{2tan^2(x)}{x} = \frac{2tanx tan x }{x } = {2 \frac{sinx}{cosx} \frac{sin }{cos } \lim_{x \to 0}[/tex] Okay this is annoying, I don't know whats wrong with my latex code, I could swear I've writen everything the same style as you. I'm not gonna try to rewrite my whole idea in latex until I figure it out. What did I do wrong here? lim x> 0 (2tan^2x)/x = [ 2 tanx (tan x) ] / x = [ 2 (sin / cos) (sin/cos) ] / x From here on out, I decided to multiply the numerator and denominator by cos/sin = [ [ 2 (sin / cos) (sin/cos) ] / x ] cos/sin = [2 (cos/sin) ] / (cos/sin) = 2 I don't see that I broke any rule of algebra or misused one so please let me know. 


#9
Oct911, 09:52 PM

P: 1,306

Can someone pleasee helppp?
[tex]\lim_{x\to 0} \frac{2tan^2x}{x}[/tex] [tex]\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}[/tex] [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}[/tex] [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}[/tex] [tex]\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}[/tex] [tex]\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}[/tex] [tex]\lim_{x\to 0}= \frac{2x}{x}[/tex] [tex]\lim_{x\to 0}= 2 [/tex] I took the time to put everything in clear latex form, I really want to figure this out. Input would be GREATLY appreciated! 


#10
Oct911, 10:13 PM

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limit of the product is the product of the limits. 


#11
Oct911, 10:46 PM

P: 1,306

[tex]\lim_{x\to 0}\frac{tan2x}{x} = \lim_{x\to 0}\frac{\frac{sinx}{cosx}\frac{sinx}{cosx}}{x}[/tex] [tex]\lim_{x\to 0}=\frac{\frac {sinx}{1} \frac{sinx}{cos^2x}}{x}[/tex] [tex]\lim_{x\to 0}=(\frac{sinx}{1}\frac{sinx}{cos^2x})\frac{1}{x}[/tex] ?? So how did you go from [tex]\lim_{x\to 0}\frac{tan2x}{x}[/tex] to [tex]\lim_{x\to 0}\frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}[/tex] Even then I don't know how to solve the problem with statement above. But I first have to know how you get to that statement. Its not just about solving the problem, its very important to learn from the problem. 


#12
Oct911, 10:55 PM

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[itex]\displaystyle \tan^2(x)=\frac{\sin(x)}{\cos(x)}\frac{\sin(x)}{ \cos(x)}=\frac{\sin(x)}{1}\frac{\sin(x)}{\cos^2(x)}[/itex]
∴ [itex]\displaystyle \frac{\tan^2(x)}{x}=\frac{\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}[/itex] 


#13
Oct911, 11:14 PM

P: 1,306

[tex]\lim_{x\to 0}=\frac{\frac {sinx}{1} \frac{sinx}{cos^2x}}{x}[/tex] [tex]\lim_{x\to 0}=(\frac{sinx}{1}\frac{sinx}{cos^2x})\frac{1}{x}[/tex] I stopped here because wouldn't that actually equal: [tex]\lim_{x\to 0}=\frac {sinx}{x} \frac{sinx}{cos^2(x)^2}[/tex] 


#14
Oct911, 11:24 PM

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Almost, but: [itex]x\,\cos(x)\ne\cos(x^2)[/itex]
That is equal to [tex]\lim_{x\to 0}\left(\frac {\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}\right)=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)[/tex] 


#15
Oct911, 11:37 PM

P: 1,306

because you can't multiply x by an angle?. What theorem shows that [tex](x) cos(x) = cos(x)[/tex] or [tex](x) cos^2 x = cos(x)[/tex] Besides that, [tex]=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)[/tex] [tex]=1 \cdot \frac{0}{1}[/tex] [tex]=0[/tex] ? *takes a deep breath* 


#16
Oct911, 11:43 PM

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Multiplication is not distributive with respect to multiplication.



#17
Oct911, 11:49 PM

P: 1,306

Thanks for your help, one final thing though; if this previous calculation was correct, could you please guide me what was wrong about the other calculation? I mean, I just don't understand what I did wrong it drives me crazy lol. How am I to get better if I don't learn from my mistakes? ^.^ [tex]\lim_{x\to 0} \frac{2tan^2x}{x}[/tex] [tex]\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}[/tex] [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}[/tex] [tex]\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}[/tex] [tex]\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}[/tex] [tex]\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}[/tex] [tex]\lim_{x\to 0}= \frac{2x}{x}[/tex] [tex]\lim_{x\to 0}= 2 [/tex] 


#18
Oct911, 11:52 PM

P: 96

Personally I would just use l'Hopital's rule.



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