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To evaluate the limit
\[ \lim _{x \rightarrow-1} \frac{x^2+5x+4}{x^2-3x-4}, \]
we first factor both the numerator and the denominator. The factoring yields:
\[ \frac{x^2+5x+4}{x^2-3x-4} = \frac{(x+4)(x+1)}{(x-4)(x+1)}. \]

Notice that both the numerator and the denominator share a common factor of \(x+1\). Upon cancelling out the common factor, we are left with:
\[ \lim _{x \rightarrow-1} \frac{x+4}{x-4}. \]

Direct substitution of \(x = -1\) into the simplified expression gives us:
\[ \frac{-1+4}{-1-4} = \frac{3}{-5} = -\frac{3}{5}. \]

Therefore, the limit
\[ \lim _{x \rightarrow-1} \frac{x^2+5x+4}{x^2-3x-4} \]
evaluates to \(-\frac{3}{5}\).
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