The problem involves finding the limit as x approaches 0 of the expression (3x - sin(3x)) / (x^2 sin(x)). The context is within trigonometric limits and properties, specifically avoiding the use of L'Hôpital's rule.
There is ongoing exploration of different approaches to the limit, with some participants suggesting that certain steps may not be valid. Others are questioning the context of the problem and whether it is appropriate given the material covered in their studies.
Participants note that the problem is from a high school exercise book and mention that they have not yet learned about power series or L'Hôpital's rule, which may affect their ability to solve the limit.
Svein said:You do remember that \lim_{z\rightarrow 0}\frac{\sin(z)}{z}=1?
That step is not valid. You cannot take the limit in the numerator only, then in the denominator. The two must be done together.songoku said:\lim_{x\rightarrow 0} {\frac{3 - \frac{\sin(3x)}{x}}{x \sin(x)}}
= \lim_{x\rightarrow 0} \frac{3 - 3}{x \sin(x)}
haruspex said:That step is not valid. You cannot take the limit in the numerator only, then in the denominator. The two must be done together.
I would expand sin() as a power series, keeping two or three terms. Not sure if that would be considered allowable in your context.
What's the context for this problem? By that, I mean where did you see this problem? The limit is apparently 4.5, but the only way I've been able to get that is by using haruspex's suggestion, in addition to using Excel to approximate the limit.songoku said:Sorry that is not allowed. No other way to solve it?
It does look like your initial approach can get you part way there.songoku said:Homework Statement
Find the value of:
lim x approaches 0 of : (3x - sin 3x) / (x2 sin x)
Homework Equations
Trigonometry identity
Limit properties
No L'hopital rule
The Attempt at a Solution
I tried changing sin 3x to -4sin3x + 3 sin x but then I stuck. Is changing sin 3x correct way to start solving the question?
Thanks
Mark44 said:What's the context for this problem? By that, I mean where did you see this problem? The limit is apparently 4.5, but the only way I've been able to get that is by using haruspex's suggestion, in addition to using Excel to approximate the limit.
SammyS said:It does look like your initial approach can get you part way there.
##\displaystyle \ \frac {3x-\sin(3x)}{x^2 \sin(x)} = \frac {3x-3\sin(x)+4\sin^3(x)}{x^2 \sin(x)} \ ##
##\displaystyle \ =\frac {3x-3\sin(x)}{x^2 \sin(x)} + \frac {4\sin^3(x)}{x^2 \sin(x)} \ ##
The limit of the second term is straight forward.
The first term remains somewhat a problem. Maybe Mark or harspex has an idea for that.
SammyS said:It does look like your initial approach can get you part way there.
##\displaystyle \ \frac {3x-\sin(3x)}{x^2 \sin(x)} = \frac {3x-3\sin(x)+4\sin^3(x)}{x^2 \sin(x)} \ ##
##\displaystyle \ =\frac {3x-3\sin(x)}{x^2 \sin(x)} + \frac {4\sin^3(x)}{x^2 \sin(x)} \ ##
The limit of the second term is straight forward.
The first term remains somewhat a problem. Maybe Mark or harspex has an idea for that.
Regarding the first term of what Sammy shows above, the only techniques that I can think of are 1) expanding the sin(3x) term (which would be ##3x - \frac{3x^3}{3!}## plus terms of degree 5 and higher), or 2) using L'Hopital's rule, which has to be applied four times.. Either way gives the result I showed in my earlier post.songoku said:From a book I use in high school. This is the question from exercise in the book. The question says: find the limit of the following, then there are a lot of limit questions, from (a) to (z). One of the question is exactly as I posted. The book doesn't cover about power series and at that point (when I saw that question), I haven't learn about L'hopital rule yet.
Maybe the question is misplaced, should not be on that part of exercise. I should cover L'hopital rule or power series first before solving that type of question.