Trig limits problem, not understanding the reasoning

[mathtwit]

Example from "Calculus" Rigdon 9th ed pg 76

Original problem:
\(\displaystyle \lim{x \to 0} \huge\frac{sin4x}{tanx}\)

Step 2:
\(\displaystyle \lim{x \to 0} \huge\frac{\frac{4 sin4x}{4x}}{ \frac{sinx}{xcosx}}\)

The problem is then solved by using the Special Trigonometric Limits theorem. (and substitution, I can post the rest if you need me to)

What's the complete justification for putting in the fractions as they are? The 4x on the bottom of the top portion doesn't seem to be reasonable (though it is mostly symmetrical), nor the xcosx on the bottom.

 

[pka]

Use these to finish.
\(\displaystyle \L
\begin{array}{l}
\lim _{x \to 0} \frac{x}{{\sin (x)}} = 1 \\
\frac{{\sin (4x)}}{{\tan (x)}} = 4\frac{{\sin (4x)}}{{4x}}\frac{x}{{\sin (x)}}\cos (x) \\
\end{array}\)

 

[mathtwit]

Thanks for the reply, my concern isn't how to finish (that's done in the book already) but why is it permissable to turn sin4x into 4sin4x/4x - it seems arbitrary and inconsistent.

I could understand 4sin4x/4 for example because it's the same as * 1.

 

[pka]

When we multiply by \(\displaystyle \frac{{4x}}{{4x}}\), there is always the danger is that it might be \(\displaystyle x=0 !\)
HOWEVER, as \(\displaystyle x \to 0\), x is never zero it is only close to zero.
Therefore, it is perfectly legal to introduce that fraction

BTW: The limit is 4.

 

[mathtwit]

Thanks again,

It appears to me that we're not multiplying sin4x times 4x/4x but actually 4/4x, this is my point of confusion.

I am probably missing some trigonometric identity that permits it to be construed this way, could you point me in the right direction for this explanation please?

 

[stapel]

It appears to me that we're not multiplying sin4x times 4x/4x but actually 4/4x....

Then where did the "x" in the numerator of the second fraction come from?

Eliz.

 

[mathtwit]

It appears to me that we're not multiplying sin4x times 4x/4x but actually 4/4x....

Then where did the "x" in the numerator of the second fraction come from?

Eliz.

I wish I knew!

I guess if I had the vocabulary handy I could go look this up - it's been too many years for me to remember everything. What (of the basics regarding fractions of fractions) should I study to better understand the logic of this problem?

PS- my idea is that if you do something to one component of it you have to do the same to the rest, like in a simple fraction. IE 1/2 * 2/2 = 2/4

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
\frac{{\sin (4x)}}{{\tan (x)}} & = & \left( {\frac{{4x}}{{4x}}} \right)\left( {\frac{{\sin (4x)}}{{\tan (x)}}} \right) \\
& = & 4\left( {\frac{{\sin (4x)}}{{4x}}} \right)\left( {\frac{x}{{\frac{{\sin (x)}}{{\cos (x)}}}}} \right) \\
& = & 4\left( {\frac{{\sin (4x)}}{{4x}}} \right)\left( {\frac{x}{{\sin (x)}}} \right)\left( {\cos (x)} \right) \\
\end{array}\)

 

[mathtwit]

I sincerely thank you all for your attempts to help me understand this but it's not getting me anywhere, and my wife is also confused by it so she can't help me either.

The utter idiot explanation, drawing no assumptions seems to be what I need. Or a relevant link to a tutorial.

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
\lim _{x \to 0} \left( {\frac{{\sin (4x)}}{{4x}}} \right) & = & 1 \\
\lim _{x \to 0} \left( {\frac{x}{{\sin (x)}}} \right) & = & 1 \\
\lim _{x \to 0} \cos (x) & = & 1 \\
\\
\lim _{x \to 0} \frac{{\sin (4x)}}{{\tan (x)}} & = & 4 \\
\end{array}\)

 

[stapel]

It appears to me that we're not multiplying sin4x times 4x/4x but actually 4/4x....

Then where did the "x" in the numerator of the second fraction come from?

I wish I knew!

It came from the "x" in the numerator of "(4x)/(4x)".

Try doing the steps yourself: Convert the tangent into a quotient of sine and cosine. Multiply the whole thing by "(4x)/(4x)". Move things around until you get the more useful form provided earlier.

Eliz.

 

[mathtwit]

4x * sin(4x) = 4 sin(4x).

That just strikes me as wrong.

 

[pka]

4x * sin(4x) = 4 sin(4x).
That just strikes me as wrong.

Well it is wrong! Which one of us said otherwise?

This is correct: (4x)sin(4x) = (4)sin(4x)(x)!

 

[stapel]

4x * sin(4x) = 4 sin(4x). That just strikes me as wrong.

It is wrong, and is not what was given to you.

If you're having trouble following the steps by looking, try doing them yourself. If you can't "see" what has been displayed, trying working out the reasoning on your own: Take "sin(4x)", and multiply it by "(4x)/(4x)". Then see if you can figure out how to get "(4)[sin(4x) / (4x)] (x)".

Eliz.

 

[mathtwit]

4x * sin(4x) = 4 sin(4x).
That just strikes me as wrong.

Well it is wrong! Which one of us said otherwise?

This is correct: (4x)sin(4x) = (4)sin(4x)(x)!

\(\displaystyle \lim{x \to 0} \huge 4x \left(\frac{sin4x}{tanx}\right)\)

Thanks for your continued patience, I am sure that I am a trying student. On making a few assumptions I cannot figure out what happened to certain elements:

Assumption 1:
\(\displaystyle \lim{x \to 0} \huge\frac{\frac{4 sin4x (where is the x)}{4x (why not 1?)}}{ \frac{sinx}{xcosx (where is the 4?)}}\)

Assumption 2:

\(\displaystyle \lim{x \to 0} \huge\frac{\frac{4 sin4x (where is the x)}{4x}}{ \frac{sinx (where is the 4x?)}{xcosx (where is the 4?)}}\)

If you could highlight where those missing elements are, and why they went where they did that would probably get the light bulb working.

 

[stapel]

In your "Assumption 1", you ask "where is the 'x'?" It is in the denominator of the denominator's fraction, [sin(x)]/[x cos(x)].

You then ask "why not 1?" Because "1" is not the argument of the sine, and you are trying to find a form which allows you to apply the referenced known limit form.

You also ask "where is the '4'?" It is up in the denominator of the numerator's fraction, [4 sin(4x)]/[4x].

Instead of going straight from the first expression to the end, try working through the steps yourself. Start with [sin(4x)]/[tan(x)], and multiply by (4x)/(4x). See where this leads. Try to rearrange [4x sin(4x)]/[4x tan(x)] into (4)[sin(4x)/(4x)][1/(xtan(x)] first.

Eliz.

 

[pka]

Come on look back through all that has been posted! PLEASE!

The problem is \(\displaystyle \L
\lim _{x \to 0} \left( {\frac{{\sin (4x)}}{{\tan (x)}}} \right).\)

IT IS NOT \(\displaystyle \L
\lim _{x \to 0} (4x)\left( {\frac{{\sin (4x)}}{{\tan (x)}}} \right)!\)

 

[mathtwit]

Thanks stapel.