A little math help

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?

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?

What happens if you use the l'Hôpital rule?

Originally posted by FabianFnas
What happens if you use the l'Hôpital rule?

there is a slight problem with that...I don't know how to use it

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)?

As Fabian said, using l'hopital gives:

2cos2x/3cos3x --> 2/3

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?

You can do it without L'hospital:

(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

Originally posted by David113
... = lim 2x/3x = lim 2/3 =

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.

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.

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....

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.😕

To answer your original question: Yes. Don't worry about the other posts. L'hopital makes things easier. But since you're just starting calc, don't worry about it,

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.

2x/3x = 2/3

No need for l'Hôpital.

Originally posted by David113
2x/3x = 2/3

No need for l'Hôpital.

2x/3x = 2/3 when x = 0?
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.

Originally posted by Jirakon
To answer your original question: Yes. Don't worry about the other posts. L'hopital makes things easier. But since you're just starting calc, don't worry about it,

Thanks Jirakon, Finally a straight answer...🙄 😀

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.

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.

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.

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.

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.

Originally posted by FabianFnas
But in the original problem x=0.

Ahh, I believe this is the misunderstanding. The question asked for the limit of the expression as x approaches 0, not the value of the expression at x=0.

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.

Perhaps, yes, I leave it with this.

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."