Limit of a trig function

[Wboyt]

Hello, I need some assistance in preparation for a Calculus 1 midterm.

The limit as x --> 0 of xcsc(3x)

Thank you in advance.

 

[Deleted member 4993]

Hello, I need some assistance in preparation for a Calculus 1 midterm.

The limit as x --> 0 of xcsc(3x)

Thank you in advance.

\(\displaystyle \lim_{x\to 0}[xcsc(3x)] \ = \ \frac{1}{3}\lim_{3x\to 0}\left [\frac{3x}{sin(3x)}\right ]\)

 

[Wboyt]

Hello, I need some assistance in preparation for a Calculus 1 midterm.

The limit as x --> 0 of xcsc(3x)

Thank you in advance.

\(\displaystyle \lim_{x\to 0}[xcsc(3x)] \ = \ \frac{1}{3}\lim_{3x\to 0}\left [\frac{3x}{sin(3x)}\right ]\)

Could you please explain how that works?

 

[Wboyt]

bump

 

[Deleted member 4993]

\(\displaystyle \lim_{x\to 0}[xcsc(3x)] \ = \ \frac{1}{3}\lim_{3x\to 0}\left [\frac{3x}{sin(3x)}\right ]\)

Could you please explain how that works?

Use the fact that:

\(\displaystyle \lim_{\theta\to 0}\left [\frac{\theta}{sin(\theta)}\right ] \ = \ 1\)

 

[Wboyt]

Yes but how do you know to pull a 1/3 out and get a 3x in the numerator?

 

[Deleted member 4993]

Yes but how do you know to pull a 1/3 out and get a 3x in the numerator?

In the denominator you have sin(3x) - I need to have 3x on the numerator - and thus......