Question

1. - / 1 points find the limit, if it exists. (if an answer does not exist, enter dne.) $lim_{x

ightarrow - 5}(|x + 5| - 7x)$ resources read it

Explanation:

Step1: Analyze the absolute - value function

When \(x\to - 5\), consider the expression inside the absolute - value \(x + 5\). When \(x\to - 5\), for the absolute - value \(|x + 5|\), when \(x\to - 5\), \(|x + 5|=-(x + 5)\) since \(x+5\lt0\) when \(x\) is approaching \(-5\) from the left and right in the neighborhood of \(x=-5\).

Step2: Substitute the absolute - value expression

Substitute \(|x + 5|=-(x + 5)\) into the limit \(\lim_{x\to - 5}(|x + 5|-7x)\), we get \(\lim_{x\to - 5}[-(x + 5)-7x]\).

Step3: Simplify the expression inside the limit

Expand and simplify \(-(x + 5)-7x=-x - 5-7x=-8x - 5\).

Step4: Evaluate the limit

Now, find \(\lim_{x\to - 5}(-8x - 5)\). Using the limit rules \(\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\) and \(\lim_{x\to a}cx = c\lim_{x\to a}x\) (\(c\) is a constant), we have \(\lim_{x\to - 5}(-8x - 5)=-8\lim_{x\to - 5}x-\lim_{x\to - 5}5\). Since \(\lim_{x\to - 5}x=-5\) and \(\lim_{x\to - 5}5 = 5\), then \(-8\times(-5)-5=40 - 5=35\).

Answer:

35