# Finding Limit of Trig Func

1. Oct 9, 2011

### Nano-Passion

1. The problem statement, all variables and given/known data
$\lim_{x\to\0} \frac{2tan^2x}{x}$

3. The attempt at a solution

$\lim_{x\to\0} \frac{2tan^2x}{x} \\ = \frac{2 tanx tanx}{x} \\ = \cfrac{2 \cfrac {sin}{cos} \cfrac {sin}{cos}$

?

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

2. Oct 9, 2011

### ehild

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

ehild

3. Oct 9, 2011

### Nano-Passion

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. Oct 9, 2011

### ehild

Your idea does not help and the result is wrong.

ehild

5. Oct 9, 2011

### ehild

$$\lim_{x \to 0} \frac{tan^2(x)}{x}=\lim_{x \to 0} \frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}$$

ehild

6. Oct 9, 2011

### LawrenceC

Do you know l'Hopital's rule?

7. Oct 9, 2011

### Nano-Passion

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?

8. Oct 9, 2011

### Nano-Passion

How so, what did I do wrong?

$$\lim_{x \to 0} \frac{2tan^2(x)}{x} = \frac{2tanx tan x }{x } = {2 \frac{sinx}{cosx} \frac{sin }{cos } \lim_{x \to 0}$$

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.

Last edited: Oct 9, 2011
9. Oct 9, 2011

### Nano-Passion

$$\lim_{x\to 0} \frac{2tan^2x}{x}$$
$$\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}$$
$$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}$$
$$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}$$
$$\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}$$
$$\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}$$
$$\lim_{x\to 0}= \frac{2x}{x}$$
$$\lim_{x\to 0}= 2$$

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: Oct 9, 2011
10. Oct 9, 2011

### SammyS

Staff Emeritus
Use the above !!!!

limit of the product is the product of the limits.

11. Oct 9, 2011

### Nano-Passion

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.

$$\lim_{x\to 0}\frac{tan2x}{x} = \lim_{x\to 0}\frac{\frac{sinx}{cosx}\frac{sinx}{cosx}}{x}$$
$$\lim_{x\to 0}=\frac{\frac {sinx}{1} \frac{sinx}{cos^2x}}{x}$$
$$\lim_{x\to 0}=(\frac{sinx}{1}\frac{sinx}{cos^2x})\frac{1}{x}$$

?? So how did you go from $$\lim_{x\to 0}\frac{tan2x}{x}$$ to
$$\lim_{x\to 0}\frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}$$

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. Oct 9, 2011

### SammyS

Staff Emeritus
$\displaystyle \tan^2(x)=\frac{\sin(x)}{\cos(x)}\frac{\sin(x)}{ \cos(x)}=\frac{\sin(x)}{1}\frac{\sin(x)}{\cos^2(x)}$

∴ $\displaystyle \frac{\tan^2(x)}{x}=\frac{\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}$

13. Oct 9, 2011

### Nano-Passion

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:
$$\lim_{x\to 0}=\frac {sinx}{x} \frac{sinx}{cos^2(x)^2}$$

14. Oct 9, 2011

### SammyS

Staff Emeritus
Almost, but:  $x\,\cos(x)\ne\cos(x^2)$

That is equal to $$\lim_{x\to 0}\left(\frac {\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}\right)=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)$$

15. Oct 9, 2011

### Nano-Passion

I suppose $x\,\cos(x)\ne\cos(x^2)$
because you can't multiply x by an angle?. What theorem shows that $$(x) cos(x) = cos(x)$$ or $$(x) cos^2 x = cos(x)$$

Besides that,

$$=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)$$
$$=1 \cdot \frac{0}{-1}$$
$$=0$$

?

*takes a deep breath*

16. Oct 9, 2011

### SammyS

Staff Emeritus
Multiplication is not distributive with respect to multiplication.

17. Oct 9, 2011

### Nano-Passion

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? ^.^

$$\lim_{x\to 0} \frac{2tan^2x}{x}$$
$$\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}$$
$$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}$$
$$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}$$
$$\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}$$
$$\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}$$
$$\lim_{x\to 0}= \frac{2x}{x}$$
$$\lim_{x\to 0}= 2$$

18. Oct 9, 2011

### McAfee

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

19. Oct 10, 2011

### Nano-Passion

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.