Limit of sin2x/sin3x | Find the Solution

[Nano-Passion]

Homework Statement

Determine the limit of

\lim_{x \to 0} \frac {sin2x}{sin3x}

Homework Equations

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

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.

 

[ehild]

Do you not what is \lim _{x \to 0} \frac{sin(x)}{x}?

ehild

 

[vela]

In your book or notes, you should find how to evaluate
\lim_{x \to 0} \frac{\sin x}{x}

 

[Nano-Passion]

Do you not what is \lim _{x \to 0} \frac{sin(x)}{x}?

ehild

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

They are asking for \lim_ {x \to 0} \frac{sin2x}{sin3x}

 

[vela]

And you see absolutely no connection to that limit and the hint?

 

[Nano-Passion]

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.

 

[Nano-Passion]

Answer is 2/3.

 

[vela]

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.

 

[Nano-Passion]

I think we've all had those moments where we fail to see what in hindsight seems so obvious.

Thank you for your help by the way:)

 

[HallsofIvy]

For a more "formal" look, let u= 2x so that you have
\lim_{x\to 0}\frac{sin(2x)}{2x}= \lim_{u\to 0}\frac{sin(u)}{u}

 

[Nano-Passion]

For a more "formal" look, let u= 2x so that you have
\lim_{x\to 0}\frac{sin(2x)}{2x}= \lim_{u\to 0}\frac{sin(u)}{u}

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