I need to find Lim (x->0) arcsin(2x)/arcsin(3x) by substitution

In summary, to find Lim (x->0) arcsin(2x)/arcsin(3x), you can use a substitution of arcsin(2x) = y to get arcsin(sin(y)) for the nominator, which is equal to y. However, for the denominator, you will have arcsin(3/2 sin(y)) which may require the use of L'hopital's rule. If you are not familiar with L'hopital's rule, you may need to find another method to solve this limit.
  • #1
Petkovsky
62
0
I need to find Lim (x->0) arcsin(2x)/arcsin(3x)

I can do a substitution
arcsin(2x) = y => 2x = sin(y)

and get arcsin(sin(y)) for the nominator, which is equal to y.
However, for the denominator i get arcsin(3/2 sin(y)) which I'm not sure what to do with.

Am I on the right path?
 
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  • #2


Since arcsin(0)=0, you have a limit of the form 0/0...so why not use L'hopital's rule?
 
  • #3


gabbagabbahey said:
Since arcsin(0)=0, you have a limit of the form 0/0...so why not use L'hopital's rule?

Well unfortunatelly i can't even tough i know it, because we haven't covered that part yet.
 

1. What is the formula for finding the limit of arcsin(2x)/arcsin(3x) as x approaches 0?

The formula for finding the limit of a function as x approaches a specific value is given by:
lim (x->a) f(x) = L
In this case, the limit of arcsin(2x)/arcsin(3x) as x approaches 0 is represented as:
lim (x->0) arcsin(2x)/arcsin(3x) = L

2. Why is substitution used to find the limit in this case?

Substitution is used to find the limit in this case because it allows us to replace the variable (x) with the specific value (0) that it is approaching. This simplifies the expression and makes it easier to evaluate the limit.

3. How do you substitute x=0 in the given expression?

To substitute x=0 in the given expression, we replace every instance of x with 0. This means that the expression becomes:
lim (x->0) arcsin(2*0)/arcsin(3*0) = lim (x->0) arcsin(0)/arcsin(0)

4. What is the value of arcsin(0)/arcsin(0)?

The value of arcsin(0)/arcsin(0) is undefined. This is because the arcsin function is not defined for the input of 0. Therefore, the limit cannot be evaluated using substitution alone. Other mathematical techniques may be used to determine the limit in this case.

5. Can substitution be used to evaluate any limit?

No, substitution cannot be used to evaluate any limit. It can only be used to evaluate limits where the variable is approaching a specific value. In some cases, substitution may not work and other techniques may be required to find the limit.

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