#lim_(x->0)(2x-4)/(5x)#?

1 Answer
Feb 2, 2018

Undefined

Explanation:

#lim_(x->0)(2x-4)/(5x)#

Divide numerator and denominator by #x#:

#(2x/x-4/x)/(5x/x)#

Cancelling:

#(2cancel(x)/cancel(x)-4/x)/(5cancel(x)/cancel(x))=(2-4/x)/5#

as: #x->0^+color(white)(888)# , #(2-4/x)/5->-oo#

as: #x->0^-color(white)(8888)# , #(2-4/x)/5->oo#

#:.#

#lim_(x->0^+)(2x-4)/(5x)!=lim_(x->0^-)(2x-4)/(5x)#

Hence:

#lim_(x->0)(2x-4)/(5x)color(white)(888)# Undefined