# Limit of trig?

• Jun 2nd 2008, 04:44 PM
Coco87
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!
• Jun 2nd 2008, 04:46 PM
Jhevon
Quote:

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
• Jun 2nd 2008, 04:52 PM
Mathstud28
Quote:

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
• Jun 2nd 2008, 04:56 PM
ThePerfectHacker
Quote:

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, $\displaystyle \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = \lim_{\theta \to 0}\frac{\tan \theta}{\theta} = 1$.
• Jun 2nd 2008, 04:59 PM
Mathstud28
Quote:

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

$\displaystyle \lim_{x\to{0}}\frac{\frac{\sin(x)}{x}}{x+\frac{\ta n(x)}{x}}=\frac{1}{1+1}=\frac{1}{2}$
• Jun 2nd 2008, 05:02 PM
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. :D

(Taylor series solution will make no sense to Coco87.)
• Jun 2nd 2008, 05:10 PM
topsquark
Quote:

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
• Jun 2nd 2008, 05:10 PM
Mathstud28
Quote:

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. :D

(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
• Jun 2nd 2008, 05:13 PM
Jhevon
Quote:

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!
• Jun 2nd 2008, 05:16 PM
Coco87
Ahh, I see now.

Thanks!

I guess I should have caught that... (Doh)
• Jun 2nd 2008, 05:16 PM
Mathstud28
Quote:

Originally Posted by Jhevon
Oh come on! you know you wanted to do it, deep down inside!

.....no........not at all...(Smirk)
• Jun 2nd 2008, 06:04 PM
Jhevon
Quote:

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!
• Jun 2nd 2008, 06:09 PM
Mathstud28
Quote:

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

Nooooo! (Brokenheart)
• Jun 2nd 2008, 06:11 PM
Jhevon
Quote:

Originally Posted by Mathstud28
Nooooo! (Brokenheart)

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!