How do you determine the limit of #(x+2)/(3x^2)# as x approaches 0-?

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Answer 1 · 2018-05-18T18:29:05

#lim_(x->0-)(x+2)/(3x^2)=oo#

I would do it this way:
Let #x=1/y# with #x<0#,

#(x+2)/(3x^2)# = #(1/y+2)/(3/y^2)#

=#(y+2y^2)/(3)# (I multiplied numerator and denominator with #y^2# to get rid of the fractions in the numerator and denominator)

We see that #lim_(y->-oo)(y+2y^2)/(3)=oo# since #lim_(y->-oo)2/3y^2=oo#

Therefore #lim_(x->0-)(x+2)/(3x^2)=oo#

This makes sense, since as #x->0-# the numerator will be closer and closer to 2, i.e.

#(x+2)/(3x^2)->2/(3x^2)-> 2/0# which has #oo# as it's limit.