If \[ f(x)= \begin{cases} \dfrac{x^2-4x-5}{x+1}, & x\neq-1,\\[6pt] k, & x=-1, \end{cases} \] is continuous at \(x=-1\), then the value of \(k\) is:
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When resolving \( \frac{0}{0} \) limits involving quadratic equations, factoring out the problematic term \( (x - c) \) is highly reliable and prevents direct substitution errors.