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
Yes its equal to 1, couldn't fit it in here. But I did just get an idea at the moment. Here is what I did. 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?
[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
I'm having trouble following your logic. Nope. My test is only on Calculus I, I think that rule doesn't get introduced till later?
How so, what did I do wrong? [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.
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!
But I replied that I don't know the logic behind that step. It completely defeats the purpose if I don't know how he got to it. [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.
[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]
That is the same exact thing that I did BUT: [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]
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]
I suppose [itex]x\,\cos(x)\ne\cos(x^2)[/itex] 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*
I'm the kind of person that feels very unsatisfied without a proof or a theorem to refer to. But I guess I'll stop pulling your rear. 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]
Well I don't learn that till calculus 2, I looked it up real quick and it looks like a pretty straightforward theorem, but problem is I don't know how to take the derivative of the trigonometric function just yet. My class started a whopping (almost) 2 weeks late relative to other classes.