Question

question determine the limit shown below in simplest form. (limlimits_{x \to -4} \frac{x^2 - 25}{x^2 - 4x - 5})

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 25\) as a difference of squares: \(x^2 - 25=(x - 5)(x + 5)\).
Factor \(x^2-4x - 5\) by finding two numbers that multiply to \(-5\) and add to \(-4\), which are \(-5\) and \(1\): \(x^2-4x - 5=(x - 5)(x+1)\).
So the expression becomes \(\lim_{x
ightarrow - 4}\frac{(x - 5)(x + 5)}{(x - 5)(x + 1)}\).

Step2: Cancel common factors

Cancel the common factor \((x - 5)\) (since \(x
ightarrow - 4\), \(x
eq5\), so we can cancel):
\(\lim_{x
ightarrow - 4}\frac{x + 5}{x + 1}\).

Step3: Substitute \(x=-4\)

Substitute \(x = - 4\) into \(\frac{x + 5}{x + 1}\):
\(\frac{-4 + 5}{-4 + 1}=\frac{1}{-3}=-\frac{1}{3}\).

Answer:

\(-\frac{1}{3}\)