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} left( square
ight)) (simplify your answer.)

Explanation:

Step1: Factor the numerator

We factor the quadratic expression \(x^2 - 3x - 4\). 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 common factor \((x + 1)\) from the numerator and the denominator.
After canceling, we get \(\lim_{x
ightarrow - 1}(x - 4)\).

Answer:

\(x - 4\)