- 13 Sep '08 18:18 / 2 editshello,

here is the question

Find the Limit of sin2x/sin3x as x approaches 0

i have it started:

start by breaking it up

sin2x*( 1/sin3x)

(3x/2x)*(sin2x)*(1/sin3x)*(2/3)

(2/3)*(sin2x/2x)*(3x/sin3x)

(2/3)*(1)*(3x/sin3x)

in the next step, am I allowed to assume that the limit of (3x/sin3x) is the recipricol of the limit of (sin3x/3x) = 1/1=1, or am i missing another manipulation? - 13 Sep '08 19:06

What happens if you use the l'Hôpital rule?*Originally posted by joe shmo***hello,**

here is the question

Find the Limit of sin2x/sin3x as x approaches 0

i have it started:

start by breaking it up

sin2x*( 1/sin3x)

(3x/2x)*(sin2x)*(1/sin3x)*(2/3)

(2/3)*(sin2x/2x)*(3x/sin3x)

(2/3)*(1)*(3x/sin3x)

in the next step, am I allowed to assume that the limit of (3x/sin3x) is the recipricol of the limit of (sin3x/3x) = 1/1=1, or am i missing another manipulation? - 13 Sep '08 20:36 / 1 edit

there is a slight problem with that...I don't know how to use it*Originally posted by FabianFnas***What happens if you use the l'Hôpital rule?**

I'm in my first few weeks of CALC 1.

However, i did give myself the benifit of the doubt and looked the rule up on Wikipedia.

It states that if f(x)/g'(x) converges, the f(x)/g(x) converges to the same limit

the problem with this is that I don't know how to differentiate g(x), which I think in this case is (1/sin3x)? - 14 Sep '08 02:49 / 1 edit

You can do it without L'hospital:*Originally posted by joe shmo***hello,**

here is the question

Find the Limit of sin2x/sin3x as x approaches 0

i have it started:

start by breaking it up

sin2x*( 1/sin3x)

(3x/2x)*(sin2x)*(1/sin3x)*(2/3)

(2/3)*(sin2x/2x)*(3x/sin3x)

(2/3)*(1)*(3x/sin3x)

in the next step, am I allowed to assume that the limit of (3x/sin3x) is the recipricol of the limit of (sin3x/3x) = 1/1=1, or am i missing another manipulation?

(all limits are as x goes to 0)

lim sin(2x)/sin(3x) =

lim sin(2x)/sin(3x) * 1 * 1 =

lim sin(2x)/sin(3x) * lim 2x/sin(2x) * lim sin(3x)/3x =

lim [sin(2x) * 2x * sin(3x)] / lim [sin(3x) * sin(2x) * 3x] =

lim 2x/3x =

lim 2/3 =

2/3 - 14 Sep '08 06:18 / 3 edits

so your saying "technically" you cant dismiss the x's as the above poster did, because of the indeterminate form 0/0? this isnt really helping me out....*Originally posted by FabianFnas***lim 2x/3x you get a "0/0" expression on which you apply the l'Hôpital rule by your spinal cord.**

But not knowing l'Hôpital rule you are in trouble again.

perhaps if you show me how the algebra for the differentation of g(x) into g'(x) works, I may come closer to understanding both the dilemma, and the solution. - 14 Sep '08 09:58

2x/3x = 2/3 when x = 0?*Originally posted by David113***2x/3x = 2/3**

No need for l'Hôpital.

Substitute x with 0 and you get 2*0 / 3*0 giving 0/0.

I haven't heard of that 0/0 equals approx 0.6667... in general.

I don't know any method to divide anything with zero, not even zero itself. For me it's a no no.

But of course, knowing your l'Hôpital rule it's obvious that 2x/3x when x=0 is exact 2/3. Spinal knowledge. - 15 Sep '08 11:39 / 1 editFabian, lim 2x/3x = 2/3 simply because the x's cancel.

L'Hopital works too, but is like using a cannon to kill an ant.

A definition of limits involves sequences.

If for any sequence a(1), a(2), a(3), ... (with a(n) not 0 for any n) converging to 0 the limit of 2a(n)/3a(n) = c for increasing n.

then the limit of 2x/3x = c for x going to 0.

Since for all sequences you choose c = 2/3 (because the a(n)'s cancel!) we have that 2x/3x converges to 0 as x goes to 0. - 15 Sep '08 15:12

Well you cannot divide the numerator and denominator by x when x=0 just to get away the x's, because then you divide with zero. You just can't do it.*Originally posted by TheMaster37***Fabian, lim 2x/3x = 2/3 simply because the x's cancel.**

L'Hopital works too, but is like using a cannon to kill an ant.

A definition of limits involves sequences.

If for any sequence a(1), a(2), a(3), ... (with a(n) not 0 for any n) converging to 0 the limit of 2a(n)/3a(n) = c for increasing n.

then the limit of 2x/3x = c for x going to 0.

S ...[text shortened]... choose c = 2/3 (because the a(n)'s cancel!) we have that 2x/3x converges to 0 as x goes to 0.

When you do it as you say, then you really use l'Hôpital rule, but you don't know it.

If I do it without refer to l'Hôpital rule in my mat exams, I get a red -1 in the corner, and a note "Don't divide by zero!" And I agree.

But of course when you solve the problem "What is lim 2x/3x when x -> 1" then of course you can just skip the x's, because then x=1, and you divide the numerator and denominator by x and the result will be 2/3.

But in the original problem x=0. - 15 Sep '08 15:29

Perhaps, yes, I leave it with this.*Originally posted by PBE6***Ahh, I believe this is the misunderstanding. The question asked for the limit as x approaches 0, no the value of the expression at x=0.**

My teacher said this: "If you solve a problem mechanically, without knowing what the underlying principles are, then you are bound to get the wrong result from time to time."