What is the limit as x approaches 0 of #cotx/lnx#?

1 Answer
Dec 11, 2014

Since #ln x# is only defined when #x>0#, we can only talk about the right-hand limit:

#lim_{x to 0^+}{cot x}/{ln x}#

Since

#{(lim_{x to 0^+}cot x=lim_{x to 0^+}{cos x}/{sin x}={1}/{0^+}=+infty),(lim_{x to 0^+}ln x=-\infty):}#,

by applying l'H#hat{"o"}#pital's Rule (#infty#/#infty#),

#=lim_{x to 0^+}{-csc^2x}/{1/x}#

by #csc x=1/{sin x}#,

#=-lim_{x to 0^+}x/{sin^2x}#

by l'H#hat{"o"}#pital's Rule (#0"/"0#),

#=-lim_{x to 0^+}1/{2sinx cosx}=-1/0^+=-infty#

The graph of #y={cot x}/{ln x}# looks like:

enter image source here

As shown above, as #x# approaches 0 from the right, the graph is shooting down toward #-infty#.