Question

Does a value of k exist such that the following limit exists? if so, find the value of k and the corresponding limit. if not, explain why not

lim x->3 2x^2 + kx -9/ x^2- 4x +3

Answer 1

k = -3

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 ...

`(2x^2-3x-9)/(x^2-4x+3)=((x-3)(2x+3))/((x-3)(x-1))=(2x+3)/(x-1)\quad x\\e3`

The limit as x→3 is 9/2.