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

$\displaystyle\lim_{x\to7}\frac{3x^2-15x+W}{x-7}$

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,

$\displaystyle\lim_{x\to7}\frac{3x^2-15x-42}{x-7}=\lim_{x\to7}(6x-15)=27$