Limit (sin(4x)/sin(6x)) as x->0

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In summary, the conversation discusses a calculus problem involving finding the limit of a trigonometric expression without using L'Hospital's rule. The answer is found by rewriting the expression and using a known limit formula. One person suggests using L'Hospital's rule, but is reminded that it is not allowed. The final answer is determined to be 2/3.
  • #1
adamjts
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Hi all,

I'm just beginning calculus and I'm having trouble figuring this one out.

I need to solve this one without using l'hospital's rule.



Homework Statement



Find

Limit (sin(4x)/sin(6x)) as x->0

Homework Equations



We know that limit (sin(x)/x) as x -> 0 equals 1



Thanks! Much appreciated!
 
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  • #2
adamjts said:
Hi all,

I'm just beginning calculus and I'm having trouble figuring this one out.

I need to solve this one without using l'hospital's rule.



Homework Statement



Find

Limit (sin(4x)/sin(6x)) as x->0

Homework Equations



We know that limit (sin(x)/x) as x -> 0 equals 1



Thanks! Much appreciated!

In order to use the limit you now, you must rewrite the expression so that you have a factor
[itex] \frac{\sin(4x)}{4x} [/itex] right? And you need to rewrite the [itex] \frac{1}{\sin(6x)} [/itex] in what form?
 
  • #3
OH. so i'd go

= limit (sin(4x)/1) * Limit (1/sin(6x))

= Limit (4sin(4x)/4) * Limit (6/6sin(6x))

= 4/6

=2/3

Yeah?
 
  • #4
adamjts said:
OH. so i'd go

= limit (sin(4x)/1) * Limit (1/sin(6x))

= Limit (4sin(4x)/4) * Limit (6/6sin(6x))

= 4/6

=2/3

Yeah?

Almost! You need to get [itex] \frac{\sin(4x)}{4x} [/itex] in order to take the limit, not
[itex] \frac{\sin(4x)}{4} [/itex]. And same thing for the sin(6x).So you need to correct one step. But your final answer is correct. Good job!
 
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  • #5
This can be done by
L HOSPITAL's rule
 
  • #6
Differentiate with respect to x on denomenator and numerator and you get the answer
 
  • #7
Ans is 2/3
 
  • #8
Midhun said:
This can be done by
L HOSPITAL's rule

The question asks to NOT use L'Hospital's rule. Which is why I suggested this approach.
 

1. What is the value of the limit of sin(4x)/sin(6x) as x approaches 0?

The limit of sin(4x)/sin(6x) as x approaches 0 is equal to 2/3.

2. How do you solve the limit of sin(4x)/sin(6x) as x->0?

To solve this limit, you can use the L'Hopital's rule and take the derivative of both the numerator and denominator. This will result in (4cos(4x))/(6cos(6x)). Plugging in x=0 will give you the answer of 2/3.

3. Can the limit of sin(4x)/sin(6x) as x approaches 0 be solved using the Squeeze Theorem?

Yes, the Squeeze Theorem can be used to solve this limit. By setting the limit as x approaches 0, we can see that it is bounded between -2/3 and 2/3. Therefore, the limit must also be equal to 2/3.

4. How does the value of the limit of sin(4x)/sin(6x) as x->0 change if the function is multiplied by a constant?

The value of the limit will remain the same regardless of the constant that is multiplied. This is because the constant will cancel out when taking the derivative using L'Hopital's rule or when using the Squeeze Theorem.

5. Is there any other method to solve the limit of sin(4x)/sin(6x) as x approaches 0?

Yes, you can also solve this limit by using the trigonometric identity sin(2x) = 2sin(x)cos(x). By substituting this into the original limit, we get (2sin(2x))/(3sin(2x)cos(x)). Canceling out the sin(2x) terms and plugging in x=0 will give us the answer of 2/3.

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