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

[adamjts]

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!

 

[nrqed]

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
\frac{\sin(4x)}{4x} right? And you need to rewrite the \frac{1}{\sin(6x)} in what form?

 

[adamjts]

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?

 

[nrqed]

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 \frac{\sin(4x)}{4x} in order to take the limit, not
\frac{\sin(4x)}{4}. And same thing for the sin(6x).So you need to correct one step. But your final answer is correct. Good job!

 

[Midhun]

This can be done by
L HOSPITAL's rule

 

[Midhun]

Differentiate with respect to x on denomenator and numerator and you get the answer

 

[Midhun]

Ans is 2/3

 

[nrqed]

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.