# Finding Limit of Trig Func

by Nano-Passion
Tags: func, limit, trig
 P: 1,306 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 Help please. Knowing me the answer is probably pretty simple. =D
 HW Helper Thanks P: 10,598 Do you know the limit x-->0 (sin(x)/x)? ehild
P: 1,306
 Quote by ehild 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?

 HW Helper Thanks P: 10,598 Finding Limit of Trig Func Your idea does not help and the result is wrong. ehild
 HW Helper Thanks P: 10,598 $$\lim_{x \to 0} \frac{tan^2(x)}{x}=\lim_{x \to 0} \frac{sin(x)}{x}\frac{sin(x)}{cos^2(x)}$$ ehild
 P: 1,195 Do you know l'Hopital's rule?
P: 1,306
 Quote by 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
I'm having trouble following your logic.

 Quote by LawrenceC 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?
P: 1,306
 Quote by ehild Your idea does not help and the result is wrong. ehild
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.
 P: 1,306 Can someone pleasee helppp? $$\lim_{x\to 0} \frac{2tan^2x}{x}$$ $$\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}$$ $$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}$$ $$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}$$ $$\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}$$ $$\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}$$ $$\lim_{x\to 0}= \frac{2x}{x}$$ $$\lim_{x\to 0}= 2$$ I took the time to put everything in clear latex form, I really want to figure this out. Input would be GREATLY appreciated!
Emeritus
HW Helper
PF Gold
P: 7,808
 Quote by 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
Use the above !!!!

limit of the product is the product of the limits.
P: 1,306
 Quote by SammyS 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.

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

?? 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.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,808 $\displaystyle \tan^2(x)=\frac{\sin(x)}{\cos(x)}\frac{\sin(x)}{ \cos(x)}=\frac{\sin(x)}{1}\frac{\sin(x)}{\cos^2(x)}$ ∴ $\displaystyle \frac{\tan^2(x)}{x}=\frac{\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}$
P: 1,306
 Quote by SammyS $\displaystyle \tan^2(x)=\frac{\sin(x)}{\cos(x)}\frac{\sin(x)}{ \cos(x)}=\frac{\sin(x)}{1}\frac{\sin(x)}{\cos^2(x)}$ ∴ $\displaystyle \frac{\tan^2(x)}{x}=\frac{\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}$
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:
$$\lim_{x\to 0}=\frac {sinx}{x} \frac{sinx}{cos^2(x)^2}$$
 Emeritus Sci Advisor HW Helper PF Gold P: 7,808 Almost, but:  $x\,\cos(x)\ne\cos(x^2)$ That is equal to $$\lim_{x\to 0}\left(\frac {\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}\right)=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)$$
P: 1,306
 Quote by SammyS Almost, but:  $x\,\cos(x)\ne\cos(x^2)$ That is equal to $$\lim_{x\to 0}\left(\frac {\sin(x)}{x}\frac{\sin(x)}{\cos^2(x)}\right)=\lim_{x\to 0}\left(\frac {\sin(x)}{x}\right)\cdot \lim_{x\to 0}\left(\frac{\sin(x)}{\cos^2(x)}\right)$$
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*
 Emeritus Sci Advisor HW Helper PF Gold P: 7,808 Multiplication is not distributive with respect to multiplication.
P: 1,306
 Quote by SammyS 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? ^.^

$$\lim_{x\to 0} \frac{2tan^2x}{x}$$
$$\lim_{x\to 0} =\frac{2 tan x (tan x)}{x}$$
$$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x}$$
$$\lim_{x\to 0}= \frac{2\frac{sin}{cos}\frac{sin}{cos}}{x} \frac{\frac{cos}{sin}}{\frac{cos}{sin}}$$
$$\lim_{x\to 0}= \frac{\frac{2cos}{sin}}{\frac{cosx}{sinx}}$$
$$\lim_{x\to 0}= \frac{2cos}{sin}\frac{sinx}{cosx}$$
$$\lim_{x\to 0}= \frac{2x}{x}$$
$$\lim_{x\to 0}= 2$$
 P: 96 Personally I would just use l'Hopital's rule.

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