Question

find the following limit or state that it does not exist.
\\( \lim \limits_{x \to -1} \frac{4(2x - 1)^2 - 36}{x + 1} \\)

simplify the given limit.
\\( \lim \limits_{x \to -1} \frac{4(2x - 1)^2 - 36}{x + 1} = \lim \limits_{x \to -1} \left( \square \
ight) \\) (simplify your answer.)

Explanation:

Step1: Factor the numerator

First, notice that the numerator \(4(2x - 1)^2 - 36\) can be factored as a difference of squares. Let's rewrite it:
\(4(2x - 1)^2 - 36 = [2(2x - 1)]^2 - 6^2\)
Using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\), where \(a = 2(2x - 1)\) and \(b = 6\), we get:
\([2(2x - 1)-6][2(2x - 1)+6]\)
Simplify each factor:
For \(2(2x - 1)-6\):
\(4x-2 - 6=4x - 8 = 4(x - 2)\)
For \(2(2x - 1)+6\):
\(4x-2 + 6=4x + 4 = 4(x + 1)\)
So the numerator factors to \(4(x - 2)\times4(x + 1)=16(x - 2)(x + 1)\) (Wait, actually, let's do it correctly. Wait, the original numerator is \(4(2x - 1)^2-36\), so factoring out 4 first:
\(4[(2x - 1)^2 - 9]\)
Now, \((2x - 1)^2 - 9\) is a difference of squares: \((2x - 1 - 3)(2x - 1 + 3)=(2x - 4)(2x + 2)=2(x - 2)\times2(x + 1)=4(x - 2)(x + 1)\)
Then multiply by the 4 we factored out earlier: \(4\times4(x - 2)(x + 1)=16(x - 2)(x + 1)\)? Wait, no, wait:
Wait, \(4[(2x - 1)^2 - 9]=4[(2x - 1)-3][(2x - 1)+3]=4(2x - 4)(2x + 2)=4\times2(x - 2)\times2(x + 1)=16(x - 2)(x + 1)\)? Wait, no, 2x - 4 is 2(x - 2), 2x + 2 is 2(x + 1), so multiplying those gives 4(x - 2)(x + 1), then multiply by 4: 16(x - 2)(x + 1). But then the denominator is \(x + 1\), so we can cancel \(x + 1\) (since \(x\to - 1\), \(x
eq - 1\) when taking the limit, so we can cancel the common factor).

Wait, maybe a better way: Let's expand the numerator first.

\(4(2x - 1)^2-36=4(4x^2-4x + 1)-36=16x^2-16x + 4 - 36=16x^2-16x - 32\)

Now, factor the numerator: \(16x^2-16x - 32 = 16(x^2 - x - 2)=16(x - 2)(x + 1)\) (since \(x^2 - x - 2=(x - 2)(x + 1)\) by factoring: looking for two numbers that multiply to -2 and add to -1, which are -2 and 1).

So the original limit is \(\lim_{x\to - 1}\frac{16(x - 2)(x + 1)}{x + 1}\)

Step2: Cancel the common factor

Since \(x\to - 1\), \(x
eq - 1\) in the limit process, so we can cancel the \(x + 1\) terms:

\(\lim_{x\to - 1}\frac{16(x - 2)(x + 1)}{x + 1}=\lim_{x\to - 1}16(x - 2)\)

Answer:

\(16(x - 2)\)