1. ## Limit of trig?

Hey,
This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
The problem is: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\ theta}}$

I know the answer is: $\displaystyle \frac{1}{2}$

I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

I get as far as: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\ theta\cos{\theta}+\sin{\theta}}$

Thanks!

2. Originally Posted by Coco87
Hey,
This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
The problem is: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\ theta}}$

I know the answer is: $\displaystyle \frac{1}{2}$

I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

I get as far as: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\ theta\cos{\theta}+\sin{\theta}}$

Thanks!
How about using L'Hopital's? Is that allowed?

Mathstud will give a series solution shortly xD

3. Originally Posted by Coco87
Hey,
This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
The problem is: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\ theta}}$

I know the answer is: $\displaystyle \frac{1}{2}$

I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

I get as far as: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\ theta\cos{\theta}+\sin{\theta}}$

Thanks!
$\displaystyle \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}$

$\displaystyle x+\tan(x)=x+\bigg[x+\frac{1}{3}x^3+\frac{2}{15}x^5+...\bigg]=2x+\frac{1}{3}x^3+\frac{2}{15}x^5\sim{2x}$

$\displaystyle \therefore\lim_{x\to{0}}\frac{sin(x)}{x+\tan(x)}=\ lim_{x\to{0}}\frac{x}{2x}=\frac{1}{2}$

Sigh..I am caugh in a stereotype

4. Originally Posted by Coco87
Hey,
This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
The problem is: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\ theta}}$

I know the answer is: $\displaystyle \frac{1}{2}$

I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

I get as far as: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\ theta\cos{\theta}+\sin{\theta}}$

Thanks!
$\displaystyle \frac{\sin \theta}{\theta + \tan \theta} = \frac{\frac{\sin \theta}{\theta}}{1 + \frac{\tan \theta}{\theta}}$.

Now use the fact, $\displaystyle \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = \lim_{\theta \to 0}\frac{\tan \theta}{\theta} = 1$.

5. Originally Posted by Coco87
Hey,
This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
The problem is: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\ theta}}$

I know the answer is: $\displaystyle \frac{1}{2}$

I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

I get as far as: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\ theta\cos{\theta}+\sin{\theta}}$

Thanks!
Alternatively,

$\displaystyle \lim_{x\to{0}}\frac{\sin(x)}{x+\tan(x)}=\lim_{x\to {0}}\frac{\frac{\sin(x)}{x}}{1+\frac{\tan(x)}{x}}$

Now since it is commonly known that $\displaystyle \lim_{x\to{0}}\frac{\sin(x)}{x}=1$
so then

$\displaystyle \lim_{x\to{0}}\frac{tan(x)}{x}=\lim_{x\to{0}}\frac {\sin(x)}{\cos(x)x}=1\cdot\lim_{x\to{0}}\frac{\sin (x)}{x}=1$

or you can also do this as rewriting it as

$\displaystyle \lim_{x\to{0}}\frac{\tan(x)-\tan(0)}{x-0}$ and seeing the defiition of the derivative

either way we ahve

$\displaystyle \lim_{x\to{0}}\frac{\frac{\sin(x)}{x}}{x+\frac{\ta n(x)}{x}}=\frac{1}{1+1}=\frac{1}{2}$

6. \displaystyle \begin{aligned}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }} {{\theta + \tan \theta }} &= \mathop {\lim }\limits_{\theta \to 0} \left\{ {\frac{{\sin \theta }} {\theta }\cdot \frac{\theta } {{\theta +\tan\theta }}} \right\}\\ &= \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }} {\theta } \cdot \mathop {\lim }\limits_{\theta\to 0} \frac{1} {{1 + \dfrac{{\tan \theta }} {\theta }}}\\ &= \frac{1} {2}.\end{aligned}

This is equal to THP's work. While I was posting he beat me to it.

(Taylor series solution will make no sense to Coco87.)

7. Originally Posted by Coco87
Hey,
This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
The problem is: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\ theta}}$

I know the answer is: $\displaystyle \frac{1}{2}$

I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

I get as far as: $\displaystyle \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\ theta\cos{\theta}+\sin{\theta}}$

Thanks!
Or you could just use L'Hopital's rule. (Ducks the flying rotten fruit.)

$\displaystyle \lim_{\theta \to 0} \frac{sin(\theta)}{\theta + tan(\theta)}$

$\displaystyle = \lim_{ \theta \to 0} \frac{cos(\theta)}{1 + sec^2(\theta)}$

$\displaystyle = \frac{1}{1 + 1} = \frac{1}{2}$

-Dan

8. Originally Posted by Krizalid
\displaystyle \begin{aligned}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }} {{\theta + \tan \theta }} &= \mathop {\lim }\limits_{\theta \to 0} \left\{ {\frac{{\sin \theta }} {\theta }\cdot \frac{\theta } {{\theta +\tan\theta }}} \right\}\\ &= \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }} {\theta } \cdot \mathop {\lim }\limits_{\theta\to 0} \frac{1} {{1 + \dfrac{{\tan \theta }} {\theta }}}\\ &= \frac{1} {2}.\end{aligned}

This is equal to THP's work. While I was posting he beat me to it.

(Taylor series solution will make no sense to Coco87.)
Yes, I know, that is why I posted an alternate solution(same as yours actually), but I had to sate Jhevon's need for power series xD

9. Originally Posted by Mathstud28
...but I had to sate Jhevon's need for power series xD
Oh come on! you know you wanted to do it, deep down inside!

10. Ahh, I see now.

Thanks!

I guess I should have caught that...

11. Originally Posted by Jhevon
Oh come on! you know you wanted to do it, deep down inside!
.....no........not at all...

12. Originally Posted by topsquark
Or you could just use L'Hopital's rule. (Ducks the flying rotten fruit.)

$\displaystyle \lim_{\theta \to 0} \frac{sin(\theta)}{\theta + tan(\theta)}$

$\displaystyle = \lim_{ \theta \to 0} \frac{cos(\theta)}{1 + sec^2(\theta)}$

$\displaystyle = \frac{1}{1 + 1} = \frac{1}{2}$

-Dan
You know what you made me realize topsquark? I think like a physicist! I've never even done physics!

13. Originally Posted by Jhevon
You know what you made me realize topsquark? I think like a physicist! I've never even done physics!
Nooooo!

14. Originally Posted by Mathstud28
Nooooo!
Yes, it's true Mathstud... i'm a physicist at heart. i know it's hard to believe, but i can't keep lying to myself anymore!