Find Value of Limit Involving Trig. Identity w/o L'hopital Rule

[songoku]

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

 

[Svein]

You do remember that $\lim_{z\rightarrow 0}\frac{\sin(z)}{z}=1$?

 

[songoku]

You do remember that $\lim_{z\rightarrow 0}\frac{\sin(z)}{z}=1$?

Yes I do but I don't know how to use it to solve this question.

Dividing all the terms by x resulting in:

$\lim_{x\rightarrow 0} {\frac{3 - \frac{\sin(3x)}{x}}{x \sin(x)}}$

$= \lim_{x\rightarrow 0} \frac{3 - 3}{x \sin(x)}$

Then stuck

 

[haruspex]

$\lim_{x\rightarrow 0} {\frac{3 - \frac{\sin(3x)}{x}}{x \sin(x)}}$

$= \lim_{x\rightarrow 0} \frac{3 - 3}{x \sin(x)}$

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.

 

[songoku]

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.

Sorry that is not allowed. No other way to solve it?

 

[Mark44]

Sorry that is not allowed. No other way to solve it?

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]

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

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 haruspex has an idea for that.

 

[songoku]

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.

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.

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.

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.
Thanks a lot for all the help

 

[Mark44]

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.

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.

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.