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13 Sep 08
2 edits
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
Originally posted by joe shmoWhat happens if you use the l'Hôpital rule?
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?
- Joined
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13 Sep 08
1 edit
Originally posted by FabianFnasthere is a slight problem with that...I don't know how to use it
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
1 edit
Originally posted by joe shmoYou can do it without L'hospital:
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
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14 Sep 08
3 edits
Originally posted by FabianFnasso 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....
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
Originally posted by David1132x/3x = 2/3 when x = 0?
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
1 edit
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.
15 Sep 08
Originally posted by TheMaster37Well 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.
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
Originally posted by PBE6Perhaps, yes, I leave it with this.
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."