Finding Limit of Trig Func

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In summary: That is the same exact thing that I did BUT:\lim_{x\to 0}=\frac{\frac {sinx}{1} \frac{sinx}{cos^2x}}{x}\lim_{x\to 0}=(\frac{sinx}{1}\frac{sinx}{cos^2x})\frac{1}{x}I stopped here because wouldn't that actually equal:\lim_{x\to 0}=\frac {sinx}{x} \frac{sinx}{cos^2(x)^2}The error is that you wrote $sinx/cos^2(x)^2$ instead of $sinx/cos^2(x)$. And
  • #1
Nano-Passion
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Homework Statement


[itex]\lim_{x\to\0} \frac{2tan^2x}{x}[/itex]

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. :(

Homework Statement


lim x--> 0 (2tan^2x)/x

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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  • #2
Do you know the limit x-->0 (sin(x)/x)?

ehild
 
  • #3
ehild said:
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?
 
  • #4
Your idea does not help and the result is wrong. ehild
 
  • #5
[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
Do you know l'Hopital's rule?
 
  • #7
ehild said:
[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.

LawrenceC said:
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?
 
  • #8
ehild said:
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 what's wrong with my latex code, I could swear I've written everything the same style as you. I'm not going to 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.
 
Last edited:
  • #9
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!
 
Last edited:
  • #10
ehild said:
[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.
 
  • #11
SammyS said:
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.
 
  • #12
[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
SammyS said:
[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]
 
  • #14
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
SammyS said:
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*
 
  • #16
Multiplication is not distributive with respect to multiplication.
 
  • #17
SammyS said:
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]
 
  • #18
Personally I would just use l'Hopital's rule.
 
  • #19
McAfee said:
Personally I would just use l'Hopital's rule.

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.
 

1) What is a limit of a trigonometric function?

A limit of a trigonometric function is the value that a function approaches as its input approaches a certain value. It is a fundamental concept in calculus and is used to evaluate the behavior of a function at a particular point or as the input approaches a certain value.

2) How do you find the limit of a trigonometric function algebraically?

The process of finding the limit of a trigonometric function algebraically involves simplifying the function and then evaluating the limit using direct substitution. If the resulting expression is undefined, other methods such as factoring, rationalizing, or using trigonometric identities can be used to simplify and evaluate the limit.

3) Can the limit of a trigonometric function exist at a discontinuity?

No, the limit of a function cannot exist at a point where the function is discontinuous. This is because the value of the limit represents the behavior of the function as the input approaches a certain value, and if the function is discontinuous at that point, its behavior cannot be determined.

4) Are there specific rules for finding the limit of trigonometric functions?

Yes, there are several rules that can be used to find the limit of a trigonometric function, such as the sum and difference rules, product and quotient rules, and the chain rule. These rules allow for the evaluation of more complex trigonometric functions by breaking them down into simpler components.

5) Why is finding the limit of trigonometric functions important?

Finding the limit of trigonometric functions is important because it allows for the understanding of the behavior of a function at a particular point or as the input approaches a certain value. It is also crucial in solving real-world problems involving trigonometric functions, and is used extensively in calculus and other branches of mathematics.

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