1. ## [SOLVED] Trigonometrc Limit

Hi guys!

Someone in the tutoring center at my school asked me to do a limit problem for their calc 1 class, and I came up with the following solution. I would like someone to come up with an alternate solution, I don't really like mine, since I applied a theorem I believe to be true but don't recall ever seeing it done. I ended up with the right answer though, but I don't know if that was a coincidence.

Here's the "theorem" I used:

Theorem: If $\lim_{x \to a} f(x) = L$ then $\lim_{x \to a} \frac {1}{f(x)} = \frac {1}{L}$. Provided $L < \infty$ and $L \neq 0$

Proof:

Assume $\lim_{x \to a } f(x) = L$.

Then $\lim_{x \to a} \frac {1}{f(x)} = \frac {\lim_{x \to a} 1}{\lim_{x \to a} f(x)}$ ..........since we are not dealing with infinite limits, the limit of the quotient is equal to the quotient of the limits (that's true right? )

Thus we have $\lim_{x \to a} \frac {1}{f(x)} = \frac {1}{L}$

QED

Now here's the question, and my solution.

Find $\lim_{ \theta \to 0} \frac {\sin \theta}{ \theta + \tan \theta}$

(Yeah, I know, you guys thought I'd have a much harder problem for you, well no, i'm only smart enough to tackle easy problems)

Solution:

$\lim_{ \theta \to 0} \frac {\sin \theta}{ \theta + \tan \theta} = \lim_{ \theta \to 0} \frac { \frac {1}{ \csc \theta}}{ \theta + \frac { \sec \theta }{ \csc \theta}}$

.................... $= \lim_{ \theta \to 0} \frac { \frac {1}{ \csc \theta}}{ \frac { \theta \csc \theta + \sec \theta}{ \csc \theta}}$

.................... $= \lim_{ \theta \to 0} \frac {1}{ \theta \csc \theta + \sec \theta}$

.................... $= \lim_{ \theta \to 0} \frac {1}{ \frac { \theta}{ \sin \theta} + \sec \theta}$

.................... $= \frac {\lim_{\theta \to 0} 1}{\lim_{\theta \to 0} \frac { \theta }{ \sin \theta} + \lim_{\theta \to 0} \sec \theta}$

Now recall that $\lim_{\theta \to 0} \frac { \sin \theta }{ \theta } = 1$.

Applying the theorem I attempted to prove above we have that $\lim_{ \theta \to 0} \frac { \theta }{ \sin \theta} = 1$ as well (This is the part I'm not sure about, I've never this limit defined as that before)

So, $\frac { \lim_{ \theta \to 0} 1}{ \lim_{ \theta \to 0} \frac { \theta }{ \sin \theta} + \lim_{ \theta \to 0} \sec \theta} = \frac {1}{1 + 1}$

Therefore $\boxed { \lim_{ \theta \to 0 } \frac {\sin \theta}{ \theta + \tan \theta} = \frac {1}{2}}$

Can anyone come up with a better and/or shorter method? (L'Hopital's is not allowed).

Thanks

2. $\displaystyle \lim_{\theta\to 0}\frac{\sin\theta}{\theta +\tan\theta}=\lim_{\theta\to 0}\frac{\frac{\sin\theta}{\theta}}{1+\frac{\tan\th eta}{\theta}}=\frac{1}{1+1}=\frac{1}{2}$
because $\displaystyle\lim_{\theta\to 0}\frac{\tan\theta}{\theta}=1$ also.

3. Originally Posted by red_dog
$\displaystyle \lim_{\theta\to 0}\frac{\sin\theta}{\theta +\tan\theta}=\lim_{\theta\to 0}\frac{\frac{\sin\theta}{\theta}}{1+\frac{\tan\th eta}{\theta}}=\frac{1}{1+1}=\frac{1}{2}$
because $\displaystyle\lim_{\theta\to 0}\frac{\tan\theta}{\theta}=1$ also.
Well, that was simple. Thanks.

$\displaystyle\lim_{\theta\to 0}\frac{\tan\theta}{\theta}=1$ is not a limit that shows up anywhere in my textbook, but it is obvious now that i see it. That was nice! I almost doubt there will be a shorter way to do it.

Now i'm a guy who likes to see several ways of doing the same problem. Does anyone else have another solution? remember, no L'Hopital Rule

4. $
\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta +\tan(\theta)}= \lim_{\theta\to 0}\frac{1}{\theta/\sin(\theta) +\tan(\theta)/\sin(\theta)}=\lim_{\theta\to 0}\frac{1}{\theta/\sin(\theta) +1/\cos(\theta)}$

............... $=\frac{1}{\lim_{\theta\to 0}[\theta/\sin(\theta) +1/\cos(\theta)]}=\frac{1}{1+1}=\frac{1}{2}
$

RonL

5. Similar to CB, you could also divide by tan.

$\lim_{{\theta}\rightarrow{0}}\frac{\frac{sin{\thet a}}{tan{\theta}}}{\frac{\theta}{tan{\theta}}+\frac {tan{\theta}}{tan{\theta}}}$

= $\lim_{{\theta}\rightarrow{0}}\frac{cos{\theta}}{\f rac{\theta}{tan{\theta}}+1}$

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

See, lotsa ways to skin this cat.

6. a different way

7. Originally Posted by galactus
Similar to CB, you could also divide by tan.

$\lim_{{\theta}\rightarrow{0}}\frac{\frac{sin{\thet a}}{tan{\theta}}}{\frac{\theta}{tan{\theta}}+\frac {tan{\theta}}{tan{\theta}}}$

= $\lim_{{\theta}\rightarrow{0}}\frac{cos{\theta}}{\f rac{\theta}{tan{\theta}}+1}$

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

See, lotsa ways to skin this cat.
yeah, there is. but like i said, the thing that was bothering me is that i wasn't sure if the "inverse" limits were true, that is, $\lim_{x \to 0} \frac { \theta }{ \sin \theta} = 1 \mbox { and } \lim_{x \to 0} \frac { \theta }{ \tan \theta} = 1$. Those limits seem to be an obvious consequence of their inverse limits, but a lot of times things seem to make sense to me when they don't actually make sense, so i wanted to be sure. i tried to avoid taking such limits which is why i went to using things like $\csc \theta \mbox { and } \sec \theta$ but eventually i just gave up

Originally Posted by curvature
a different way
again, this method relies on the fact that the limit of the inverse of a function is the inverse of the limit of the original function, a concept i wasn't absolutely sure of until now. but thanks

8. Originally Posted by Jhevon
Here's the "theorem" I used:

Theorem: If $\lim_{x \to a} f(x) = L$ then $\lim_{x \to a} \frac {1}{f(x)} = \frac {1}{L}$. Provided $L < \infty$ and $L \neq 0$

Proof:

Assume $\lim_{x \to a } f(x) = L$.

Then $\lim_{x \to a} \frac {1}{f(x)} = \frac {\lim_{x \to a} 1}{\lim_{x \to a} f(x)}$ ..........since we are not dealing with infinite limits, the limit of the quotient is equal to the quotient of the limits (that's true right? )

Thanks
It seem to me that you have used what you are proving.
A serious proof should involve epsilon-delta.

9. Originally Posted by curvature
It seem to me that you have used what you are proving.
i didn't know if my proof was valid, i had this overwhelming suspicion that i was over looking something.

That being said, i would like you guys to comment on my proof, how was it? would the theorem be true if $L \mbox { is } \infty$ or $L \mbox { is } 0$? Of course we would define "division by infinity" as 0 and "division by 0" as infinity, we're dealing with limits so we can do that, since L may never actually attain the value 0 and no number can actually attain infinity

A serious proof should involve epsilon-delta.
Haha, i completely agree, and i could fit them in my proof if i wanted to, but i didn't think it was completely necessary and i was already babbling a lot. but your right, if there is one thing i have learned in the proof class i took last semester it's that there's always an epsilon and he's always greater than zero, otherwise, the universe will explode. even if you don't see him, he's there in the background, as he is in my proof. delta is okay, he shows up every now and then, but epsilon is always there!

10. No need to use $\delta-\epsilon$. Because there is an equivalent sequencial definiton*.

Let $f(x)$ be defined on an open interval $I - \{a\}$ containing $a$ such that $f(x)\not = 0$ on $I - \{a\}$. And $\lim_{x\to a}f(x)=L$. Then that means for any sequence $(x_n)$ converging to $a$ we have $\lim f(x_n)=L$. So then $\lim \frac{1}{f(x_n)} = \frac{1}{L}$ by the simple property of sequence limits. Hence, by the sequencial definition $\lim_{x\to a}\frac{1}{f(x)} = \frac{1}{L}$.

Remark: Jhevon did a minor mistake. He said $L\not =0$ which is good, but the above proof shows that we need that fact $f(x) \not =0$ for some open interval containing $a$ except possible at $a$.

*)Let $f(x)$ be defined on $I-\{a\}=S$ we define $\lim_{x\to a}f(x) = L$ iff there exists a non-empty $J\subset S$ open interval such that for all $x_n\in J$ converging to $a$ we have $f(x_n)$ converge to $L$.

11. Originally Posted by Jhevon
Then $\lim_{x \to a} \frac {1}{f(x)} = \frac {\lim_{x \to a} 1}{\lim_{x \to a} f(x)}$
The problem is that you have used the above un-proved rule, which is stronger than what you are proving.

12. Originally Posted by ThePerfectHacker
... we need that fact $f(x) \not =0$ for some open interval containing $a$ except possible at $a$.
yeah, i see that. maybe that was one of the things that were bugging me

Originally Posted by curvature
The problem is that you have used the above un-proved rule, which is stronger than what you are proving.
i believe that is one of the well-known theorems of limits, you generally don't have t prove those. i was proving a specific property of limits and assumed all the other common properties were known to the reader (hopefully i knew what i was talking about as well ).

13. Originally Posted by ThePerfectHacker
No need to use $\delta-\epsilon$. Because there is an equivalent sequencial definiton*.

Let $f(x)$ be defined on an open interval $I - \{a\}$ containing $a$ such that $f(x)\not = 0$ on $I - \{a\}$. And $\lim_{x\to a}f(x)=L$. Then that means for any sequence $(x_n)$ converging to $a$ we have $\lim f(x_n)=L$. So then $\lim \frac{1}{f(x_n)} = \frac{1}{L}$ by the simple property of sequence limits. Hence, by the sequencial definition $\lim_{x\to a}\frac{1}{f(x)} = \frac{1}{L}$.
The use of $\delta-\epsilon$ is standard in most calculus books. Your method is right, but too serious for non-math major guys i'm afraid.

14. Originally Posted by curvature
The use of $\delta-\epsilon$ is standard in most calculus books. Your method is right, but too serious for non-math major guys i'm afraid.
i am a math major... and i am familiar with the definition used by TPH, and it is, i believe, in the calculus textbook used at our school

15. Originally Posted by curvature
The problem is that you have used the above un-proved rule, which is stronger than what you are proving.
I agree with Jhevon there is no need to prove all limits using $\delta - \epsilon$ given that we are using known theorems (as long as the theorem is used properly). The same with other areas in mathematics.

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