lim x->0 sin2x/sin3x

[vjekobalas]

I saw this old problem and answer (copy below) on some forum but
still didn't understand the solution i.e. if we substitute 1 for sin2x/2x and sin3x/3x
we are left with the 2x/3x - how is that supposed to be = 1 /what is the
thinking behind this ? Is it that as x->0, 2x and 3x are approx the same and
therefore 2x/3x approx 1 ?


******************
Proof: special limit is that lim as x->0 (sinA/A) = 1

So sin2x/(2x) = 1

To get sin (2x)/2x, multiply both sides of the fraction by 2x. Similarly, multiply both sides by 3x. You get:

lim x -> 0 of (2x)*[sin2x/(2x)]* 3x/{3x*sin3x}
which equals 2x/3x = 2/3

 

[pka]

I saw this old problem and answer (copy below) on some forum but
still didn't understand the solution i.e. if we substitute 1 for sin2x/2x and sin3x/3x
we are left with the 2x/3x - how is that supposed to be = 1 /what is the
thinking behind this ? Is it that as x->0, 2x and 3x are approx the same and
therefore 2x/3x approx 1 ?


******************
Proof: special limit is that lim as x->0 (sinA/A) = 1

So sin2x/(2x) = 1

To get sin (2x)/2x, multiply both sides of the fraction by 2x. Similarly, multiply both sides by 3x. You get:

lim x -> 0 of (2x)*[sin2x/(2x)]* 3x/{3x*sin3x}
which equals 2x/3x = 2/3

Anyone should see: \(\displaystyle \dfrac{\sin(2x)}{\sin(3x)}=\dfrac{2}{3}\dfrac{\sin(2x)}{2x}\dfrac{3x}{\sin(3x)}\)

 

[stapel]

I saw this old problem and answer (copy below) on some forum but still didn't understand the solution i.e. if we substitute 1 for sin2x/2x and sin3x/3x we are left with the 2x/3x - how is that supposed to be = 1 /what is the thinking behind this ?

Since you've posted this to an "Algebra" category, I'll guess that you've never taken calculus. You'll learn "the thinking behind this" when you take the first semester of calculus. (By the way, the "=1" thing refers to the sine parts, not the x parts.)

 

[pka]

lim(f(x)/g(x))= lim (f'(x)/g'(x))

Is this true?
\(\displaystyle \Large{\displaystyle{\lim _{x \to 0}}\frac{{x - 1}}{{\sqrt x }} = {\lim _{x \to 0}}\frac{{\frac{{d\left( {x - 1} \right)}}{{dx}}}}{{\frac{{d\left( {\sqrt x } \right)}}{{dx}}}}}\)

 

[Prove It]

lim(f(x)/g(x))= lim (f'(x)/g'(x))

As long as it's a 0/0 or infinity/infinity indeterminate form.

Is this true?
\(\displaystyle \Large{\displaystyle{\lim _{x \to 0}}\frac{{x - 1}}{{\sqrt x }} = {\lim _{x \to 0}}\frac{{\frac{{d\left( {x - 1} \right)}}{{dx}}}}{{\frac{{d\left( {\sqrt x } \right)}}{{dx}}}}}\)

No, as it's not one of those indeterminate forms.

 

[vjekobalas]

Looks like my question wasn't very clear or there's some interpretations going on
pka - where do you get 2/3 from, from 2x/3x - that is my question ?
bestellen - thanks, but that doesn't help me and certainly doesn't answer the question
stapel - that's clear about sinx/x - I was asking about the 2x/3x

 

[pka]

No, as it's not one of those indeterminate forms.

Why not let the student discover that?

 

[Ishuda]

I saw this old problem and answer (copy below) on some forum but
still didn't understand the solution i.e. if we substitute 1 for sin2x/2x and sin3x/3x
we are left with the 2x/3x - how is that supposed to be = 1 /what is the
thinking behind this ? Is it that as x->0, 2x and 3x are approx the same and
therefore 2x/3x approx 1 ?


******************
Proof: special limit is that lim as x->0 (sinA/A) = 1

So sin2x/(2x) = 1

To get sin (2x)/2x, multiply both sides of the fraction by 2x. Similarly, multiply both sides by 3x. You get:

lim x -> 0 of (2x)*[sin2x/(2x)]* 3x/{3x*sin3x}
which equals 2x/3x = 2/3

As you say, you are left with \(\displaystyle \frac{2\, x}{3\, x}\) which is equal to \(\displaystyle \frac{2}{3}\), not 1 [just cancel the x in both numerator and denominator].

 

[stapel]

I saw this old problem and answer (copy below) on some forum...

If they really said that the limit of (2x)/(3x) was 1, then they were mistaken. It would be helpful if you provided a link to that original source. For instance, this discussion comes to the standard conclusion (not "1", as in whatever you'd seen). Thank you!