Question

How do you evaluate the limit `abs(x^2-9)/abs(x-3)` as x approaches `3`?

Answer 1

`=6`

Explanation:

Substituting `3` in the given expression, to find its limit, results to
`" "`
an indeterminate solution :

`lim_(x->3)abs(x^2-9)/abs(x-3)`

`=(abs(3^2-9))/(abs(3-3))`

`=0/0` `" "` Indeterminate solution.
`""`

Factorizing `" "x^2-9" "` is the ideal way to find the solution.

`x^2-9=x^2-3^2=(x-3)(x+3)`
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`lim_(x->3^-)abs(x^2-9)/abs(x-3)=lim_(x->3^-)abs((x-3)(x+3))/abs(x-3)`
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#=lim_(x->3^-)(abs(x-3)abs(x+3))/abs(x-3)" " =lim_(x->3^-)((3-x)(x+3))/(3-x)#
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`=lim_(x->3^-)(cancel((3-x))(x+3))/cancel((3-x))=lim_(x->3^-)3+x=3+3=6`
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`" "`
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`lim_(x->3^+)abs(x^2-9)/abs(x-3)=lim_(x->3^+)abs((x-3)(x+3))/abs(x-3)`
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#=lim_(x->3^+)(abs(x-3)abs(x+3))/abs(x-3)" " =lim_(x->3^+)((x-3)(x+3))/(x-3)#
`" "`
`=lim_(x->3^+)(cancel((x-3))(x+3))/cancel((x-3))=lim_(x->3^+)x+3=3+3=6`
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Therefore,
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`lim_(x->3)abs(x^2-9)/abs(x-3)=lim_(x->3^-)abs(x^2-9)/abs(x-3)=lim_(x->3^+)abs(x^2-9)/abs(x-3)=6`