Limit of sin2x/sin3x | Find the Solution

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In summary, the conversation discusses finding the limit of \lim_{x \to 0} \frac {sin2x}{sin3x} and the use of a hint to evaluate it. The conversation also includes a reminder of the limit \lim _{x \to 0} \frac{sin(x)}{x} and the solution of the original limit being 2/3. One member also suggests using a substitution to evaluate the limit in a more formal manner.
  • #1
Nano-Passion
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Homework Statement



Determine the limit of

[itex] \lim_{x \to 0} \frac {sin2x}{sin3x}[/itex]

Homework Equations



Hint: Find [itex]\lim_{x\to 0} (\frac{2 sin 2x}{2x}) (\frac{3x}{3 sin 3x})[/itex]

The Attempt at a Solution



I've been blankly staring at it not knowing where to start. I think the only thing that the hint manages to do is to confuse me.

Any help on helping me to start it? I don't understand the hint.

Thanks in advance.
 
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  • #2
Do you not what is [tex] \lim _{x \to 0} \frac{sin(x)}{x}[/tex]?

ehild
 
  • #3
In your book or notes, you should find how to evaluate
[tex]\lim_{x \to 0} \frac{\sin x}{x}[/tex]
 
  • #4
ehild said:
Do you not what is [tex] \lim _{x \to 0} \frac{sin(x)}{x}[/tex]?

ehild

Yes they are equal to one, but they aren't asking for [tex] \lim_{x \to 0} \frac {sinx}{x}[/tex]

They are asking for [tex]\lim_ {x \to 0} \frac{sin2x}{sin3x}[/tex]
 
  • #5
And you see absolutely no connection to that limit and the hint?
 
  • #6
vela said:
And you see absolutely no connection to that limit and the hint?

Ohhhhhh !

Its funny how someone can say so little yet help so much hehe. Thanks, I got it now.
 
  • #7
Answer is 2/3.
 
  • #8
Nano-Passion said:
Its funny how someone can say so little yet help so much hehe. Thanks, I got it now.
I think we've all had those moments where we fail to see what in hindsight seems so obvious. :wink:
 
  • #9
vela said:
I think we've all had those moments where we fail to see what in hindsight seems so obvious. :wink:

Thank you for your help by the way:)
 
  • #10
For a more "formal" look, let u= 2x so that you have
[tex]\lim_{x\to 0}\frac{sin(2x)}{2x}= \lim_{u\to 0}\frac{sin(u)}{u}[/tex]
 
  • #11
HallsofIvy said:
For a more "formal" look, let u= 2x so that you have
[tex]\lim_{x\to 0}\frac{sin(2x)}{2x}= \lim_{u\to 0}\frac{sin(u)}{u}[/tex]

Hey halls, thanks for the suggestion. Will keep in mind in the future. ^.^
 

FAQ: Limit of sin2x/sin3x | Find the Solution

What is the limit of sin2x/sin3x as x approaches 0?

The limit of sin2x/sin3x as x approaches 0 is equal to 2/3.

How do you find the limit of sin2x/sin3x?

To find the limit of sin2x/sin3x, you can use the limit laws and apply the fact that sinx/x approaches 1 as x approaches 0. This gives us 2/3 as the limit.

What is the domain of sin2x/sin3x?

The domain of sin2x/sin3x is all real numbers except for multiples of pi/3, as this would result in a denominator of 0.

Can you use L'Hospital's rule to find the limit of sin2x/sin3x?

Yes, you can use L'Hospital's rule to find the limit of sin2x/sin3x. This results in the same answer of 2/3 as using the limit laws.

What is the limit of sin2x/sin3x as x approaches infinity?

The limit of sin2x/sin3x as x approaches infinity does not exist, as the function oscillates between -1 and 1 indefinitely.

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