Question

find $\lim \limits_{x \to c^+} f(x)$ and $\lim \limits_{x \to c^-} f(x)$ for the given function and value of $c$.
$f(x) = (x + 8)\frac{|x + 7|}{x + 7}, c = -7$
$\lim \limits_{x \to -7^+} (x + 8)\frac{|x + 7|}{x + 7} = 1$ (simplify your answer.)
$\lim \limits_{x \to -7^-} (x + 8)\frac{|x + 7|}{x + 7} = -1$ (simplify your answer.)

Explanation:

Step1: Analyze the absolute value for \(x \to -7^+\)

When \(x \to -7^+\), \(x + 7>0\), so \(|x + 7|=x + 7\). Then the function becomes \(f(x)=(x + 8)\frac{x + 7}{x + 7}=x + 8\) (for \(x
eq - 7\)).
Now, find the limit as \(x\to - 7^+\): \(\lim_{x\to - 7^+}(x + 8)=-7 + 8 = 1\).

Step2: Analyze the absolute value for \(x \to -7^-\)

When \(x \to -7^-\), \(x + 7<0\), so \(|x + 7|=-(x + 7)\). Then the function becomes \(f(x)=(x + 8)\frac{-(x + 7)}{x + 7}=-(x + 8)\) (for \(x
eq - 7\)).
Now, find the limit as \(x\to - 7^-\): \(\lim_{x\to - 7^-}-(x + 8)=-(-7 + 8)=-1\).

Answer:

\(\lim_{x\to - 7^+}(x + 8)\frac{|x + 7|}{x + 7}=1\), \(\lim_{x\to - 7^-}(x + 8)\frac{|x + 7|}{x + 7}=-1\)