Since this limit is 0/0 so we can use L'Hopital's rule

`lim_(x->1) ln(x)/(sin(5pix)) = lim_(x->1) (1/x)/(5picos(5pix))`

This limit is not 0/0 and can be evaluated `cos(5pi(1)) = -1`

`lim_(x->1) f(x) = (1/1)/(5pi(-1)) = -1/(5pi)`

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now