Question

Evaluate the limit analytically:

Answer 1

\[\LARGE \lim_{\theta \rightarrow 0}~\frac{\sin8\theta}{4\theta^2}*\sin5(\theta)\]

Answer 2

You can change it to the product of two standard limits.

Do you know \[\lim_{x \rightarrow 0} \frac{ \sin(x) }{ x }\]

Answer 3

That would be 1 if I'm not mistaken

Answer 4

\[\lim_{\theta \rightarrow0}\frac{ \sin 8\theta }{ 8 \theta }\lim_{\theta \rightarrow 0}\frac{\sin 5 \theta }{5 \theta } \times \frac{ 8\times5 }{4 }\]

Answer 5

Yes! But if you look well, you can see that\[\lim_{\theta \rightarrow 0}\frac{ \sin(8\theta) }{ 4\theta ^{2} }=\lim_{\theta \rightarrow 0}\frac{ \sin(8\theta) }{ 4\theta }\times\frac{ \sin(5\theta) }{ \theta }\]

Answer 6

@surjithayer: it is better to explain than to do everything yourself imo.

Answer 7

Now if you look at the two fractions, you will see that they are beginning to look like the standard limit. The next step will make them even more alike...

Answer 8

if there is any doubt she can ask me.

Answer 9

sin(x)/x = x as x--> 0, standard small angle limit. You can do the derivation of this limit from scratch with the arguments in this problem, or use trig identities to put this in the form where it is a product of C(sin(x)/x). This is a very volatile limit, the sin function oscillates wildly near x=0, and even more so when there is a higher constant multiplied by the argument. I would use trig identities

Answer 10

Because the numerator of the first fraction has 8 theta, we write that in the denominator as well.
The second fraction: 5 theta in the numerator, so 5 theta also below:\[\lim_{\theta \rightarrow 0}\frac{ \sin(8\theta) }{ 8\theta }\times \lim_{\theta \rightarrow 0}\frac{ \sin5\theta }{ 5\theta }\]Of course, we had to cheat a little here, because we borrowed some numbers to get what we want.
Therefore, we have to compensate for this: we began with only the number 4 in the denominator and ended with 8 and 5 as factors, giving 40.
If we multiply with 10 everything will be fine again:

Answer 11

\[\lim_{\theta \rightarrow 0}\frac{ \sin(8\theta) }{ 4\theta^{2} }\times \sin(5\theta)=10\times\lim_{\theta \rightarrow 0}\frac{ \sin(8\theta) }{ 8\theta }\times \lim_{\theta \rightarrow 0}\frac{ \sin(5\theta) }{ 5\theta }\]

Answer 12

There is only one problem left: we have 8 theta in the fraction, but in the limit, it says that theta goes to 0. The limits we have made are still not equal to the standard limit \[\lim_{number \rightarrow 0}\frac{ \sin(number) }{ number }=1\]So we have to change variables:
say phi = 8theta, then, if theta goes to 0, phi will also go to 0.
Doing the same for 5theta, we get:

\[(original limit)=10 \times \lim_{\phi \rightarrow 0}\frac{ \sin(\phi) }{ \phi }\times \lim_{\alpha \rightarrow 0}\frac{ \sin(\alpha) }{ \alpha }=10 \times 1 \times 1 = 10\]The last step (changing the variables to get the exact standard limit is often considered "just extra work, because you already got it" , but it is necessary I'm afraid...

So a lot of work, but just using a very well-known standard limit. There is certainly no reason to use "trig identities" (which ones?) as @kbomeisl suggested!

Answer 13

do you know where the small angle limit comes from? The answer is trig identities. When we do more rigorous problems, like normalizing quantum wave functions, we need to calculate trigonometric limits like the small angle approximation you rely on from scratch rather than just citing it, sometimes imaginary angles are involved, etc. You can just modify the approach of the proof of the small angle approximation by putting other arguments than x in there, like 4x, for example.

Answer 14

So much information to take in at once xD
Thanks for for your help guys, really appreciate it :)

Answer 15

YW!

Answer 16

@kbomeisl: I do not want to be rude, so if you think I am, please accept my apologies.
I only think that there is nothing more here than standard limits.
The standard limit of sin(x)/x has been proven to be 1 very rigourously. From that point on, other limits looking like it can be evaluated as well, by using the value of this standard limit.

Answer 17

Yeah, you are right ZeHanz, that is the best approach to solve this problem. I'm just wired to solve it the other way.

Answer 18

@kbomeisl: Most people here are not busy normalizing quantum wave functions, I think ;)