How do you find the limit of #(ln(x)/sin(πx))# as x approaches 1?

Question

15694 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-07T23:15:53

#= - 1/pi#

this is meant as another L'Hopital thing as it is #0/0# indeterminate

so with #lim_(x to 1) (ln(x)/sin(πx))#

we can walk right in by saying that

# = lim_(x to 1) ((1/x)/(pi cos (πx)))#

and these are all continuous through the limit!

# =1/pi lim_(x to 1) 1/x * (1)/( cos (πx))#

so we can separate them out
# =1/pi lim_(x to 1) 1/x * lim_(x to 1) ((1)/( cos (πx)))#

# =1/pi * 1 * (-1)#

#= - 1/pi#