Question

How do you find the limit of `(x^2-3x+2)/(x^3-4x)` as x approaches 2 from the right, as x approaches -2 from the right, as x approaches 0 from the left, and as x approaches 1 from the right?

Answer 1

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.