Question

question
determine the limit shown below in simplest form.
(limlimits_{x \to -7} \frac{x^2 - 1}{x^2 + 6x - 7})
answer attempt 1 out of 2
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dne

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(x^2 - 1\) using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\). So, \(x^2 - 1=(x - 1)(x + 1)\).

Next, factor the denominator \(x^2 + 6x - 7\). We need two numbers that multiply to \(-7\) and add to \(6\). Those numbers are \(7\) and \(-1\). So, \(x^2 + 6x - 7=(x + 7)(x - 1)\).

Now the limit becomes:
\[
\lim_{x
ightarrow - 7}\frac{(x - 1)(x + 1)}{(x + 7)(x - 1)}
\]

Step2: Cancel common factors

We can cancel out the common factor \((x - 1)\) (assuming \(x
eq1\), which is fine here since we are taking the limit as \(x
ightarrow - 7\), not \(x = 1\)):
\[
\lim_{x
ightarrow - 7}\frac{x + 1}{x + 7}
\]

Step3: Substitute \(x=-7\) (Check for indeterminate form)

Now, substitute \(x=-7\) into \(\frac{x + 1}{x + 7}\). The denominator becomes \(-7 + 7 = 0\) and the numerator becomes \(-7+1=-6\). So we have \(\frac{-6}{0}\), which means the limit does not exist (DNE) because we have a non - zero number divided by zero.

Answer:

DNE