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Finding Limit of Trig Func

by Nano-Passion
Tags: func, limit, trig
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Nano-Passion
#1
Oct9-11, 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
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ehild
#2
Oct9-11, 01:02 AM
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Do you know the limit x-->0 (sin(x)/x)?

ehild
Nano-Passion
#3
Oct9-11, 01:12 AM
P: 1,306
Quote Quote by ehild View Post
Do you know the limit x-->0 (sin(x)/x)?

ehild
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?

ehild
#4
Oct9-11, 01:39 AM
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Finding Limit of Trig Func

Your idea does not help and the result is wrong.


ehild
ehild
#5
Oct9-11, 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
LawrenceC
#6
Oct9-11, 07:32 AM
P: 1,195
Do you know l'Hopital's rule?
Nano-Passion
#7
Oct9-11, 02:17 PM
P: 1,306
Quote Quote by ehild View Post
[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.

Quote Quote by LawrenceC View Post
Do you know l'Hopital's rule?
Nope. My test is only on Calculus I, I think that rule doesn't get introduced till later?
Nano-Passion
#8
Oct9-11, 02:28 PM
P: 1,306
Quote Quote by ehild View Post
Your idea does not help and the result is wrong.


ehild
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.
Nano-Passion
#9
Oct9-11, 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!
SammyS
#10
Oct9-11, 10:13 PM
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Quote Quote by ehild View Post
[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
Use the above !!!!

limit of the product is the product of the limits.
Nano-Passion
#11
Oct9-11, 10:46 PM
P: 1,306
Quote Quote by SammyS View Post
Use the above !!!!

limit of the product is the product of the limits.
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.
SammyS
#12
Oct9-11, 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]
Nano-Passion
#13
Oct9-11, 11:14 PM
P: 1,306
Quote Quote by SammyS View Post
[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]
SammyS
#14
Oct9-11, 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]
Nano-Passion
#15
Oct9-11, 11:37 PM
P: 1,306
Quote Quote by SammyS View Post
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*
SammyS
#16
Oct9-11, 11:43 PM
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Multiplication is not distributive with respect to multiplication.
Nano-Passion
#17
Oct9-11, 11:49 PM
P: 1,306
Quote Quote by SammyS View Post
Multiplication is not distributive with respect to multiplication.
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]
McAfee
#18
Oct9-11, 11:52 PM
P: 96
Personally I would just use l'Hopital's rule.


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