What is the limit of # (sinx)(ln4x)# as x approaches 0?

Question

9074 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-09T08:01:54

0

when written as

#lim_(x to 0) (ln 4x)/(1/(sin x))# this is #oo/oo# indeterminate so we can use L'Hopital's Rule

#= lim_(x to 0) (1/x)/(- cscx cot x)#

#= - lim_(x to 0) sin x tan x * (1/x)#

using well known limit:# lim_(z to 0) (sin z)/z = 1#

#= - lim_(x to 0) (sin x)/x lim_(x to 0) tan x #

#= - 1 * 0 = 0#