The Limit of a Trigonometric Function

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Homework Help Overview

The discussion revolves around evaluating the limit of a trigonometric function as x approaches 0, specifically lim x-> 0 ( sin(5x) - 3x ) / ( 3x ). Participants are exploring the application of limit properties and trigonometric identities in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to manipulate the limit expression using known limits and algebraic rearrangements. There is a focus on the relationship between sin(x) and x as x approaches 0, and how this applies to the given limit. Questions about the correctness of their approaches and results are also raised.

Discussion Status

Some participants have provided feedback on the original poster's approach, noting errors in the algebraic manipulation. There is an ongoing exploration of the limit's value, with references to discrepancies between personal calculations and results from WolframAlpha. Additional questions about related limits are being posed, indicating a broader inquiry into the properties of trigonometric functions.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of assistance they can provide to one another. There is a noted frustration with the use of LaTeX for mathematical expressions, which may affect clarity in communication.

Jambr
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Homework Statement



lim x-> 0 ( sin(5x) - 3x ) / ( 3x )

Homework Equations



(a - b) / c = (a / c) - b

The Attempt at a Solution



(1/3) lim x->0 ( sin(5x) / x ) - 3x
= (1/3) lim x->0 ( sin(5x) / x )( 5 / 5) - 3x
= (5/3) lim x->0 ( sin(5x) / 5x ) - 3x
= (5/3) * 1 - 3x
= (5/3) - 3x
= (5/3)

I use WolframAlpha to check whether or not my answers are correct. WolframAlpha and I are having a disagreement (even though I am most likely wrong) about this problem. I say 5/3, it says 2/3. Can someone please explain where I went wrong?

Sorry, I tried using LaTeX but I just got annoyed.
 
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You can start with (a - b) / c = (a / c) - b. That's not right. (a-b)/c=(a/c)-(b/c).
 
Jambr said:

Homework Statement



lim x-> 0 ( sin(5x) - 3x ) / ( 3x )

Homework Equations



(a - b) / c = (a / c) - b

The Attempt at a Solution



(1/3) lim x->0 ( sin(5x) / x ) - 3x
= (1/3) lim x->0 ( sin(5x) / x )( 5 / 5) - 3
Here is your error.
\frac{sin(5x)- 3x}{3x}= \frac{sin(5x)}{3x}-\frac{3x}{3x}= \frac{sin(5x)}{3x}- 1
= \frac{5}{3}\frac{sin(5x)}{5x}- 1

= (5/3) lim x->0 ( sin(5x) / 5x ) - 3x
= (5/3) * 1 - 3x
= (5/3) - 3x
= (5/3)

I use WolframAlpha to check whether or not my answers are correct. WolframAlpha and I are having a disagreement (even though I am most likely wrong) about this problem. I say 5/3, it says 2/3. Can someone please explain where I went wrong?

Sorry, I tried using LaTeX but I just got annoyed.
 
Thank you, I had tried that already but I did it by removing the 3 from the bottom first off, which was wrong.

Another question: As we all know sin x / x = 1, is it also true at sin^2 x / x^2 = 1?
 
Jambr said:
Thank you, I had tried that already but I did it by removing the 3 from the bottom first off, which was wrong.

Another question: As we all know sin x / x = 1, is it also true at sin^2 x / x^2 = 1?

If you mean limit as x->0, sure. Both limits are 1.
 

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