Math Homework: Limit of sin3x/x+3x^2

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (sin(3x))/(x + 3x^2) as x approaches 0. Participants are exploring the mathematical concepts related to limits, particularly in the context of trigonometric functions and their behavior near zero.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of L'Hopital's Rule and the limit properties of sine functions. There are questions about rewriting the limit to facilitate evaluation, and some participants express differing opinions on the best approach to take.

Discussion Status

The conversation includes various perspectives on the use of L'Hopital's Rule versus other methods, such as the limit property of sine functions. Some participants suggest that rewriting the limit could simplify the evaluation, while others emphasize the validity of different approaches without reaching a consensus on the preferred method.

Contextual Notes

There are indications of confusion regarding the application of limit properties and the conditions under which L'Hopital's Rule is applicable. Participants are also clarifying the relationship between sine and linear approximations near zero.

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does this seem right?
lim (sin3x)/(x+3x^2)
x->0

=(sin3x)/[3x^2(1/3x+1)]
=1/[(1/3(0)+1)(sin3(0))]
=undefined
 
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sin(3x) = 0
x-> 0
 
do you know l'Hopital's rule?
 
Can you rewrite your equation in order to use:

[tex]\lim_{x \to 0}\frac{\sin(3x)}{3x}=1[/tex]?
 
Galileo said:
Can you rewrite your equation in order to use:

[tex]\lim_{x \to 0}\frac{\sin(3x)}{3x}=1[/tex]?
Don't you mean [tex]\lim_{x \to 0}\frac{3 \sin(x)}{3x}=1[/tex]?

Or are you just taking advantage of the fact, that near zero, sin x = x?

does this seem right?
lim (sin3x)/(x+3x^2)
x->0
L'Hopital's Rule is certainly the easiest way to solve this. If you haven't learned it yet, check the table of contents and find it. Some books teach it very early on and some wait to teach it until later on. It's easy enough to learn that I'd learn it early even if the course waits until later.

For your equation, it's simply the derivative of the top divided by the derivative of the bottom. You can do that whenever your limit winds up 0/0, or infinity/infinity.

In other words:

[tex]\lim_{x \to 0}\frac{\sin(3x)}{x+3x^2}=\lim_{x \to 0}\frac{3 \cos(3x)}{1+6x}[/tex]?
 
Last edited:
BobG said:
Don't you mean [tex]\lim_{x \to 0}\frac{3 \sin(x)}{3x}=1[/tex]

No, Galileo meant [tex]\lim_{x\to 0}\frac{sin(3x)}{3x}= 1[/tex] just like he said.

BobG said:
Or are you just taking advantage of the fact, that near zero, sin x = x?

Sort of, though I wouldn't say it like that. sin x is NOT equal to x "near zero". They are only equal AT 0. What you meant to say, I am sure, is that sin x is close to x for x close to 0 and the become close as x goes to 0: [tex]\lim_{x \to 0}\frac{sin x}{x}= 1[/tex] and [tex]\frac{sin(3x)}{x}= \frac{3 sin(3x)}{3x}= 3\frac{sin(y)}{y}[/tex] where y= 3x. In other words, [tex]\lim_{x \to 0}\frac{sin(3x)}{x}= \lim_{x \to 0}3\frac{sin 3x}{3x}= 3\lim_{y \to 0}\frac{sin(y)}{y}= 3[/tex].

In my opinion, it is far simpler to use Galileo's idea than to use L'Hopital.
 
Last edited by a moderator:
Yes, it is. L'Hopital's result is too strong for this problem. Although it will of course get you the answer quite readily, it's a lot harder to prove than the things Galileo's method needs :wink:

Nevertheless, if he did know l'Hopital's rule, it would be the obvious path~
 

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