Question

problem. 13 : evaluate the limit using the appropriate limit law(s). (if an answer does not exist, enter dne.)
\\(\lim\limits_{x \to 5} \frac{-2x + 10}{|x - 5|} =\\)
problem. 14 : evaluate the limit using the appropriate limit law(s). (if an answer does not exist, enter dne.)
\\(\lim\limits_{x \to -2} \frac{-6|x| + 12}{3x + 6} =\\)
problem. 15 : if you know that \\(\lim\limits_{x \to -1} f(x) = -3\\) and \\(\lim\limits_{x \to 0} g(x) = 3\\), then evaluate the following limit:
\\(\lim\limits_{x \to 0} f(x - 1)g(x) =\\)
problem. 16 : if you know that \\(\lim\limits_{x \to 4} f(x) = 3\\) and \\(\lim\limits_{x \to 0} g(x) = -5\\), then evaluate the following limit:
\\(\lim\limits_{x \to 0} f(x + 4) + g(x) =\\)

Explanation:

Problem 13

Step1: Analyze left and right limits

For \( \lim_{x \to 5} \frac{-2x + 10}{|x - 5|} \), we consider left - hand limit (\(x\to5^-\)) and right - hand limit (\(x\to5^+\)) separately.
First, factor the numerator: \(-2x + 10=-2(x - 5)\).
For \(x\to5^+\), \(x - 5>0\), so \(|x - 5|=x - 5\). Then the function becomes \(\frac{-2(x - 5)}{x - 5}=-2\) (for \(x
eq5\)). So \(\lim_{x\to5^+}\frac{-2x + 10}{|x - 5|}=\lim_{x\to5^+}\frac{-2(x - 5)}{x - 5}=-2\).
For \(x\to5^-\), \(x - 5<0\), so \(|x - 5|=-(x - 5)\). Then the function becomes \(\frac{-2(x - 5)}{-(x - 5)} = 2\) (for \(x
eq5\)). So \(\lim_{x\to5^-}\frac{-2x + 10}{|x - 5|}=\lim_{x\to5^-}\frac{-2(x - 5)}{-(x - 5)}=2\).

Step2: Compare left and right limits

Since \(\lim_{x\to5^-}\frac{-2x + 10}{|x - 5|}=2\) and \(\lim_{x\to5^+}\frac{-2x + 10}{|x - 5|}=-2\), and \(2
eq - 2\), the two - sided limit does not exist.

Step1: Simplify the absolute value for \(x\to - 2\)

For \(x\to - 2\), \(x<0\), so \(|x|=-x\). The function \(\frac{-6|x|+12}{3x + 6}\) becomes \(\frac{-6(-x)+12}{3x + 6}=\frac{6x + 12}{3x + 6}\).

Step2: Factor and simplify

Factor the numerator and denominator: \(6x + 12 = 6(x + 2)\), \(3x+6 = 3(x + 2)\). Then the function (for \(x
eq - 2\)) is \(\frac{6(x + 2)}{3(x + 2)} = 2\).

Step3: Evaluate the limit

\(\lim_{x\to - 2}\frac{-6|x|+12}{3x + 6}=\lim_{x\to - 2}\frac{6(x + 2)}{3(x + 2)}=\lim_{x\to - 2}2 = 2\).

Step1: Analyze \(\lim_{x\to0}f(x - 1)\)

We know that \(\lim_{x\to - 1}f(x)=-3\). Let \(t=x - 1\), when \(x\to0\), \(t\to - 1\). So \(\lim_{x\to0}f(x - 1)=\lim_{t\to - 1}f(t)=-3\).

Step2: Use the product rule for limits

The product rule for limits states that if \(\lim_{x\to a}u(x)\) and \(\lim_{x\to a}v(x)\) exist, then \(\lim_{x\to a}[u(x)v(x)]=\lim_{x\to a}u(x)\cdot\lim_{x\to a}v(x)\).
Here, \(u(x)=f(x - 1)\) and \(v(x)=g(x)\), \(a = 0\). We know that \(\lim_{x\to0}f(x - 1)=-3\) and \(\lim_{x\to0}g(x)=3\).
So \(\lim_{x\to0}f(x - 1)g(x)=\lim_{x\to0}f(x - 1)\cdot\lim_{x\to0}g(x)=(-3)\times3=-9\).

Answer:

DNE

Problem 14