Answer:
k = -3
Step-by-step explanation:
The expression evaluates at x=3 to ...
(2·3² +3k -9)/(3² -4·3 +3) = (9+3k)/0
In order for the limit to exist, there must be a "hole" at x=3. That will only be the case when ...
9 +3k = 0
k = -9/3 = -3
The value of k must be -3 for the limit to exist.
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When k=-3, the expression factors as ...
[tex]\dfrac{2x^2-3x-9}{x^2-4x+3}=\dfrac{(x-3)(2x+3)}{(x-3)(x-1)}=\dfrac{2x+3}{x-1}\quad x\ne3[/tex]
The limit as x→3 is 9/2.
