# Homework Help: Limits with trig functions manipulations

1. Feb 13, 2006

### Jeff Ford

I'm pretty useless at trig manipulations, but I'm trying to learn

According to my text, the key to limits with trig functions is to get them into this form

$$\lim_ {t\rightarrow 0} \frac{\sin x}{x} = 1$$

I've been doing alright with that, but I'm stuck on this one
$$\lim_ {t\rightarrow 0} \frac{\tan 6t}{\sin 2t}$$

I'm not sure what else to turn this into besides
$$\lim_ {t\rightarrow 0} \frac{\sin 6t}{\sin 2t \cos 6t}$$

A poke in the right direction would be appreciated.

Thanks
Jeff

Last edited: Feb 13, 2006
2. Feb 13, 2006

### quasar987

You could use the identity

$$tg2x=2tgx/(1-tg^2x)$$

3. Feb 13, 2006

### HallsofIvy

Take a look at
$$\frac{n}{m}\frac{sin 6t}{nt}\frac{mt}{sin 2t}\frac{1}{cos 6t}$$

I'll let you do the work of deciding how to choose n and m correctly.

4. Feb 13, 2006

### Jeff Ford

I do not teach a calculus class, I take a calculus class :tongue:

5. Feb 13, 2006

### Jeff Ford

$$\lim_ {t\rightarrow 0} \frac{\sin 6t}{\sin 2t \cos 6t} (\frac{12t}{12t}) = \lim_ {t\rightarrow 0} \frac{6}{2} (\frac{\sin 6t}{6t}) (\frac {2t}{\sin 2t}) (\frac{1}{\cos 6t})$$

I know the limit here for each term except $$(\frac {2t}{\sin 2t})$$

Straight substitution leaves this undeffined. Is there a standard for working out limits in this form?

Last edited: Feb 13, 2006
6. Feb 13, 2006

### assyrian_77

Is your limit $$x\rightarrow 0$$ or $$t\rightarrow 0$$? Anyway, I seem to remember that even
$$\lim_{x\rightarrow 0} \frac{x}{\sin{x}}=1$$
but I am far from certain.

7. Feb 13, 2006

### Jeff Ford

It's $t\rightarrow 0$. I've made the change in the earlier posts.

It would seem that if $$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$ then $$\lim_{x\rightarrow 0} \frac{x}{\sin x} = 1$$, but I'm wary of making that assumption. Can someone clarify if I am correct?
If so, the answer to my original question is 3, which matches up with the back of the text.

Thanks

Jeff

8. Feb 13, 2006

### assyrian_77

I seem to remember this being the case. However I was not certain so I did a quick check in Mathematica and it gives me the same answers. The question is; how reliable is Mathematica?

9. Feb 13, 2006

### arildno

Do you agree that your wariness hinges upon the following:
Given that lim f =L (non-zero), then lim 1/f =1/L ?

10. Feb 13, 2006

### Jeff Ford

It appears to be true. I can't think of a counter example.

11. Feb 13, 2006

### arildno

That's not good enough! You should prove it!
Here's one way:
1. assume that the limit of f at x=0 is L (non-zero)
Since L is non-zero, there must be a non-zero neighbourhood about x=0 where f is always non-zero (otherwise, L could not be f's limit value at x=0, since at arbitrarily close points, there would be function values differing from L at least of magnitude L)

2. Regard the difference (within the domain existing by virtue of 1.):
$$|\frac{1}{f}-\frac{1}{L}|=\frac{1}{|f||L|}|f-L|$$

try to concoct an epsilon-delta proof to show that the statement holds.

12. Feb 13, 2006

### Jeff Ford

I'm still in Calc I and haven't done much more than the most basic epsilon-delta proofs. I'd love to learn how, but I'm uncertain as to where to begin.

Do I start with $$|\frac{1}{f}-\frac{1}{L}| < \epsilon$$ and $$\frac{1}{|f||L|}|f-L| <\epsilon$$ and try to find $$\delta > |x|$$ that works for both cases?

13. Feb 13, 2006

### arildno

The first thing to remember about the epsilon-delta business is not to bother overmuch about the epsilon!
What IS in general important, is to bound your expression with a function of delta that is easily seen to go to zero as delta goes to zero.

Now, in this case there are two salient features:
1. We need to estimate the |f| in the denominator in a clever manner.
2. We have the difference |f-L| to take care of.

1. We want to find an upper bound for $\frac{1}{|f||L|}|f-L|$
Hence, we should first find a LOWER bound for |f|, since it appears in the denominator.
Since we know that the limit of f is L, there will exist a $\delta_{1}$ so that for all x's inside that delta,$$|f|\geq\frac{|L|}{2}$$
(the half is chosen for simplicity, the validity of the inequality is straightforward to see).

2. Since f goes to L, then for any $\epsilon[/tex] there will exist a [itex]\delta_{2}$ so that $|f-L|\leq\frac{L^{2}\epsilon}{2}[/tex] (This particular choice is just cosmetics, in order to end up with a nice expression.) 3. Thus, choosing [itex]\delta=minimum(\delta_{1},\delta_{2})$
we get, for x's inside the delta-band:
$$\frac{1}{|L||f|}|f-L|<\frac{2}{L^{2}}\frac{L^{2}\epsilon}{2}=\epsilon$$
Voila!

14. Feb 13, 2006

### Jeff Ford

Clever logic. It took me awhile, but I think I understand what you did. Just for reference, is this something I should be learning at the Calc I level, or does this usually come up in something like an Analysis class?

15. Feb 13, 2006

### daveb

Since you are in Calc 1 you can use L'Hospital's Rule to show that the limit exists and is also 1.

16. Feb 13, 2006

### arildno

And how do you prove in the first place that we have:
$$\frac{d}{dx}\sin(x)=\cos(x)$$ ?

17. Feb 13, 2006

### Jeff Ford

That's about a chapter ahead of where I am now. I read a few sections ahead of the class, but not quite that far.

18. Feb 13, 2006

### daveb

Ah, I didn't realize you hadn't covered derivatives yet.

19. Feb 13, 2006

### Jeff Ford

I can follow the proof of this one almost all the way to the end, staring from the limit definition of the derivative. I just get a little stuck at the last step. I can break it down to terms that are almost all defined when $h = 0$ but I get two terms in that step that are $$\lim_{h\rightarrow 0} \frac{\sin(h)}{h}$$

I know this equals one, but I'm not sure how to prove this logically. The proof in my text is only done geometrically.

20. Feb 13, 2006

### TD

Without using derivatives, you can show this since for small values of x (we're letting x going to 0), we have that:

$$\begin{array}{l} \sin x \le x \le \tan x \\ \frac{{\sin x}}{{\sin x}} \le \frac{x}{{\sin x}} \le \frac{{\tan x}}{{\sin x}} \\ 1 \le \frac{x}{{\sin x}} \le \frac{1}{{\cos x}} \\ \end{array}$$

And since $\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x}} = 1$, we've now 'squeezed' our function between two which have limit 1.