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

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The discussion revolves around evaluating the limit of (sin(3x))/(x + 3x^2) as x approaches 0. Participants suggest using L'Hopital's Rule, which involves taking the derivatives of the numerator and denominator when encountering a 0/0 form. There is a debate on whether to apply L'Hopital's Rule or to utilize the limit property that sin(x)/x approaches 1 as x approaches 0. Some argue that using the limit property is simpler and more intuitive than L'Hopital's Rule for this specific problem. Ultimately, both methods lead to the conclusion that the limit evaluates to 3.
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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:

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

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

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:

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

No, Galileo meant \lim_{x\to 0}\frac{sin(3x)}{3x}= 1 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: \lim_{x \to 0}\frac{sin x}{x}= 1 and \frac{sin(3x)}{x}= \frac{3 sin(3x)}{3x}= 3\frac{sin(y)}{y} where y= 3x. In other words, \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.

In my opinion, it is far simpler to use Galileo's idea than to use L'Hopital.
 
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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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