trig l'hospitals

[Dorian Gray]

Greetings Mathematicians,

I was wondering if somebody could please help me with a limit problem using l'hospital's rule with trigonometry. I attached a copy of my work. I am not sure how to address this problem. Any suggestions are always appreciated!
Should I try this?

 

[lookagain]

Greetings Mathematicians,

I was wondering if somebody could please help me with a limit problem using l'hospital's rule with trigonometry. I attached a copy of my work. I am not sure how to address this problem. Any suggestions are always appreciated!
View attachment 1925 Should I try this? View attachment 1924

Dorian Gray,


\(\displaystyle \dfrac{sin(6x)}{sin(2x)} \ \ne \ sin(4x)\)



It is true that \(\displaystyle \dfrac{sin^6(x)}{sin^2(x)} \ = \ sin^4(x).\)


Edit:

In your second step (meaning the step right after the original problem),
you have the form 0/0.


So, take the derivative of the numerator (recommending the product rule), * *
and take the derivative of the denominator.

Look at the limit of that resulting quotient.

See if the new quotient stops being an indeterminate form at that point.



** the fraction being \(\displaystyle \dfrac{cos(2x)sin(6x)}{sin(2x)}\)

 

[Dorian Gray]

thank you!

Thank you very much! I apologize for getting back so late. I solved it your way, and then a friend also showed me a way how to use the Sin (x)/x rule and such to solve it. ( I prefered your way)