How do you find the limit of #(2x-1)/(abs(2x^3 - x^2))# as x is approaching 0.5 from the negative side?

Question

9563 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-06T03:31:00

Factor, then use definition and properties of the absolute value.

#(2x-1)/abs(2x^3-x^2)=(2x-1)/(abs(x^2)abs(2x-1))=1/x^2(2x-1)/abs(2x-1)#

As #xrarr1/2# (from either side), #1/x^2rarr4#

For #x<0.5#, we have #2x<1# and #2x-1<0#.

Hence, for #x<0.5#, we get

#abs(2x-1)=-(2x-1)# and

#(2x-1)/abs(2x-1)=(2x-1)/-(2x-1)=-1#.

Therefore, as #xrarr1/2# from the negative side, #(2x-1)/abs(2x^3-x^2)rarr(4)(-1)=-4#.