​Compute ​

​$$\lim _{x \rightarrow 4} \frac{\left(x^2-7 x+12\right)}{x-4}$$​​

Correct option is C

​$$\begin{aligned}&\text{First, try substituting } x = 4 \text{ directly into the expression:} \\&\frac{4^2 - 7 \cdot 4 + 12}{4 - 4} = \frac{16 - 28 + 12}{0} = \frac{0}{0} \\&\text{This is an indeterminate form, so we need to simplify the expression further.} \\\\&\textbf{Factor the Numerator} \\&\text{Factor the quadratic expression in the numerator:} \\&x^2 - 7x + 12 = (x - 3)(x - 4) \\&\text{So, the limit becomes:} \\&\lim_{x \to 4} \frac{(x - 3)(x - 4)}{x - 4} \\\\&\textbf{Simplify the Expression} \\&\text{Cancel the common factor } (x - 4) \text{ in the numerator and denominator (valid since } x \ne 4\text{):} \\&\lim_{x \to 4} (x - 3) \\\\&\textbf{Evaluate the Simplified Limit} \\&\text{Now, substitute } x = 4 \text{ into the simplified expression:} \\&4 - 3 = 1\end{aligned}$$​​