Limits with trig functions manipulations

[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

 

[quasar987]

You could use the identity

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

I tought I read in another thread that you teach a calculus class. Did I misread?

 

[HallsofIvy]

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

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

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

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

A poke in the right direction would be appreciated.

Thanks
Jeff

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.

 

[Jeff Ford]

You could use the identity

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

I tought I read in another thread that you teach a calculus class. Did I misread?

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

 

[Jeff Ford]

Ok, I've made some progress

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

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?

 

[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.

 

[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

 

[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?

 

[arildno]

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

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

What do you know about this statement?

 

[Jeff Ford]

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

 

[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.

 

[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?

 

[arildno]

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?

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

 

[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?

 

[daveb]

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

 

[arildno]

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

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

 

[Jeff Ford]

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

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.

 

[daveb]

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

 

[Jeff Ford]

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

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.

 

[TD]

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

$$\begin{array}{l}<br /> \sin x \le x \le \tan x \\ <br /> \frac{{\sin x}}{{\sin x}} \le \frac{x}{{\sin x}} \le \frac{{\tan x}}{{\sin x}} \\ <br /> 1 \le \frac{x}{{\sin x}} \le \frac{1}{{\cos x}} \\ <br /> \end{array}$$

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.

 

[Jeff Ford]

Ah, the squeeze theorem. I hadn't had a chance to make use of that yet. Thanks!

 

[TD]

Ah, the squeeze theorem. I hadn't had a chance to make use of that yet. Thanks!

You're welcome

 

[arildno]

Another, geometric "proof" is the following:
1. Draw a unit circle
2. Draw a line through the centre with angle x (in radians!) to another line through the centre (say, the horizontal).
3. Consider the areas of the 3 figures:
a) the triangle with hypotenuse 1, base cos(x) and height sin(x)
(area: 1/2*sin(x)cos(x))
b) the circle sector indicated, area x/2
c) the right-angled triangle with base 1 and height tan(x)
area: tan(x)/2

Evidently, we have a<b<c, which can be rewritten as,by dividing throughout with sin(x)/2:
$$cos(x)<\frac{x}{\sin(x)}<\frac{1}{\cos(x)}$$

the value of the ratio is squeezed into 1 as x goes to zero..

 

[Jeff Ford]

The explanation I got from my instructor last night was

$$\lim_{x\rightarrow 0} \frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow 0} f(x)}{\lim_{x\rightarrow 0} g(x)}$$

Therefore

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

 

[TD]

The explanation I got from my instructor last night was

$$\lim_{x\rightarrow 0} \frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow 0} f(x)}{\lim_{x\rightarrow 0} g(x)}$$

Note that this only holds if both limits in numerator and denumerator exist.

 

[Jeff Ford]

True, I should have made that clarification. Thanks.