Question

find the following limit or state that it does not exist
(limlimits_{x \to -1} \frac{x^2 - 3x - 4}{x + 1})

simplify the given limit
(limlimits_{x \to -1} \frac{x^2 - 3x - 4}{x + 1} = limlimits_{x \to -1} (x - 4)) (simplify your answer)

evaluate the limit, if possible. select the correct choice below and, if necessary, fill in the answer box to complete your choice

(\bigcirc) a (limlimits_{x \to -1} \frac{x^2 - 3x - 4}{x + 1} = square) (type an exact answer)

(\bigcirc) b. the limit does not exist

Explanation:

Step1: Factor the numerator

We factor \(x^2 - 3x - 4\). We need two numbers that multiply to \(-4\) and add to \(-3\). Those numbers are \(-4\) and \(1\). So, \(x^2 - 3x - 4=(x - 4)(x + 1)\).
The limit becomes \(\lim_{x
ightarrow - 1}\frac{(x - 4)(x + 1)}{x + 1}\).

Step2: Cancel the common factor

Since \(x
ightarrow - 1\) but \(x
eq - 1\) (we are taking the limit, not evaluating at \(x=-1\)), we can cancel the \((x + 1)\) terms. So we get \(\lim_{x
ightarrow - 1}(x - 4)\).

Step3: Evaluate the limit

Now we substitute \(x=-1\) into \(x - 4\). So, \(-1-4=-5\).

Answer:

A. \(\lim\limits_{x
ightarrow - 1}\frac{x^{2}-3x - 4}{x + 1}=\boxed{-5}\)