Unique at Only One Point!

Notice that if we evaluate this limit without giving regards to $k$ , it will be $\lim_{x\to3} \dfrac{4(9)+10k-6}{18-15-3} = \frac{30+10k}{0}$ In this regard, if the denominator is zero but numerator is not zero, this limit will not exist (as $\frac{x}{0}$ for $x \neq 0$ is not determined). If the numerator is zero, we can conclude it to be indeterminate form of $\frac{0}{0}$ which may has limit by transformation to other form that can be determinate (whether the transformation will give another indeterminate form or not). This suggests that $30 + 10k = 0 \Rightarrow k = -3$ is a possible solution.

Substituting $k=-3$ gives us the limit,

$\begin{aligned} \lim_{x\to3} \dfrac{4x^2 - 3x + 7(-3) - 6}{2x^2 - 5x - 3} &=& \lim_{x\to3} \dfrac{4x^2 - 3x -27}{2x^2 - 5x - 3} \\ &=& \lim_{x\to3} \dfrac{(x-3) (4x+9)}{(2x+1)(x-3) } , \qquad \text{Factor the quadratic expressions} \\ &=& \lim_{x\to3} \dfrac{\cancel{(x-3)} (4x+9)}{(2x+1)\cancel{(x-3)} } \\ &=& \lim_{x\to3} \dfrac{4x+9}{2x+1} \\ &=& \dfrac{4(3)+ 9}{2(3) + 1} = \dfrac{21}7 = 3 \end{aligned}$

which is a finite number, thus this result tells us that the limit does indeed exists when $k = -3$ .

Footnote : This tells us that when $k = -3$ , the limit exists and is equal to $3$ ; and when $k\ne -3$ , the limit fails to exist (diverge).

$\begin{aligned} \lim_{x \to 3} \frac{4x^2+kx+7k-6}{2x^2-5x-3} & = \lim_{x \to 3} \frac{4x^2+kx+7k-6}{(2x+1)(x-3)} \end{aligned}$

For the expression to have a limit, $4x^2+kx+7k-6$ must have $(x-3)$ as a factor or has the form:

$\begin{aligned} 4x^2+kx+7k-6 & = (x-3)(ax+b) \\ & = ax^2 + (b-3a)x - 3b \end{aligned}$

Equating the coefficients:

$\begin{cases} a = 4 \\ b - 3a = k & \Rightarrow b = k + 12 \\ 3b = 6-7k & \Rightarrow 3(k+12) = 6 - 7k & \Rightarrow 10k = -30 & \Rightarrow k = \boxed{-3} \end{cases}$