Question

Find the numerical value for W such that l'Hospital's Rule applies to the following limit and then compute the value, L, of that limit.

lim as x approaches 7 (((3x^2)-15x+W)/(x-7))=L

Answer 1

`\displaystyle\lim_(x\to7)(3x^2-15x+W)/(x-7)`

For the limit to exist,
`x=7` needs to be a removable discontinuity. This means the numerator needs to have a factor of
`x-7`.

The polynomial remainder theorem says that a polynomial
`p(x)` has a factor
`x-r` if
`p(r)=0`. In this case,
`p(x)=3x^2-15x+W` and
`r=7`. You have
`p(7)=42+W`, and for this to be exactly 0, you require that
`W=-42`.

Now, the numerator approaches 0, and so by L'Hopital's rule,


`\displaystyle\lim_(x\to7)(3x^2-15x-42)/(x-7)=\lim_(x\to7)(6x-15)=27`