Question

1. $limlimits_{x \to -1} \frac{(2x - 1)^2 - 9}{x + 1}$

Explanation:

Step1: Factor the numerator

First, we recognize that the numerator \((2x - 1)^2 - 9\) is a difference of squares, since \(9 = 3^2\). The formula for the difference of squares is \(a^2 - b^2=(a - b)(a + b)\). Here, \(a = 2x - 1\) and \(b = 3\). So we can factor the numerator as follows:
\[

$$\begin{align*} (2x - 1)^2 - 9&=(2x - 1 - 3)(2x - 1 + 3)\\ &=(2x - 4)(2x + 2) \end{align*}$$

\]
We can further factor out a 2 from each of these binomials:
\[

$$\begin{align*} (2x - 4)(2x + 2)&=2(x - 2)\times2(x + 1)\\ &=4(x - 2)(x + 1) \end{align*}$$

\]

Step2: Simplify the function

Now we substitute the factored form of the numerator back into the original function. The original function is \(\frac{(2x - 1)^2 - 9}{x + 1}\), and after factoring the numerator, we have:
\[
\frac{4(x - 2)(x + 1)}{x + 1}
\]
Since \(x\to - 1\) but \(x
eq - 1\) (because we are taking a limit, we are looking at the behavior near \(x=-1\), not at \(x = - 1\) itself), we can cancel out the common factor of \(x + 1\) in the numerator and the denominator. This leaves us with the function \(4(x - 2)\).

Step3: Evaluate the limit

Now we can find the limit as \(x\to - 1\) of the simplified function \(4(x - 2)\). We substitute \(x=-1\) into this function:
\[
4(-1 - 2)=4\times(-3)=-12
\]

Answer:

\(-12\)