Start by factoring the expression,
#{x^2-3x+2}/{x^3-4x}={(x-2)(x-1)}/{x(x+2)(x-2)}={(x-1)}/{x(x+2)}#
Now start taking the required limits.
#lim_{x \rarr0^-}[{(x-1)}/{x(x+2)}]#
#=lim_{x \rarr0^-}[{(x-1)}/{(x+2)}]*lim_{x \rarr0^-}[1/x]#
#={(0-1)}/{(0+2)}*(-\infty)=-infty#
#lim_{x \rarr0^-}[1/x]=-infty# because we are approaching zero from the negative side of the number line.
For the next limit,
#lim_{x \rarr-2^+}[{(x-1)}/{x(x+2)}]#
#=lim_{x \rarr-2^+}[{(x-1)}/{x}]*lim_{x \rarr-2^+}[1/{(x+2)}]#
#={-3}/{-2}*(+infty)=+infty#
#lim_{x \rarr-2^+}[1/{(x+2)}]=+infty# because as #(x\rarr-2^+)#, #(x+2)# gets very small, but stays positive.
For the next limit,
#lim_{x \rarr 1^+}[{(x-1)}/{x(x+2)}]=lim_{x \rarr1^+}[{1}/{x(x+2)}]lim_{x \rarr1^+}[(x-1)]#
#=1/{1*3}*0=0#
For the last limit,
#lim_{x \rarr2^+}[ {(x-1)}/{x(x+2)}]=3/{2*4}=3/8#
Even though the denominator of the original expression went to zero at #x=2#, the limit is still finite because the numerator went to zero just as quickly. Recall, at the start we were able to cross out the #(x-2)# in the denominator (which was causing the singularity) with another #(x-2)# factor in the numerator.
