Question

find $\lim_{x \to c^+} f(x)$ and $\lim_{x \to c^-} f(x)$ for the given function and value of $c$. $f(x) = (x + 11) \frac{|x + 6|}{x + 6}$, $c = -6$
$(x + 11) \frac{|x + 6|}{x + 6} = \square$ (simplify your answer.)

Explanation:

Step1: Analyze the absolute value function

For \(|x + 6|\), when \(x\to - 6^{+}\) (approaching \(-6\) from the right), \(x+6>0\), so \(|x + 6|=x + 6\). When \(x\to - 6^{-}\) (approaching \(-6\) from the left), \(x + 6<0\), so \(|x+6|=-(x + 6)\).

Step2: Find \(\lim_{x\to - 6^{+}}f(x)\)

Substitute \(|x + 6|=x + 6\) (since \(x\to - 6^{+}\), \(x+6>0\)) into \(f(x)=(x + 11)\frac{|x + 6|}{x + 6}\). Then \(f(x)=(x + 11)\frac{x + 6}{x + 6}\). The \(\frac{x + 6}{x + 6}\) (for \(x
eq - 6\)) simplifies to \(1\). So \(\lim_{x\to - 6^{+}}f(x)=\lim_{x\to - 6^{+}}(x + 11)\). Substitute \(x=-6\) into \(x + 11\), we get \(-6 + 11 = 5\).

Step3: Find \(\lim_{x\to - 6^{-}}f(x)\)

Substitute \(|x + 6|=-(x + 6)\) (since \(x\to - 6^{-}\), \(x + 6<0\)) into \(f(x)=(x + 11)\frac{|x + 6|}{x + 6}\). Then \(f(x)=(x + 11)\frac{-(x + 6)}{x + 6}\). The \(\frac{-(x + 6)}{x + 6}\) (for \(x
eq - 6\)) simplifies to \(- 1\). So \(\lim_{x\to - 6^{-}}f(x)=\lim_{x\to - 6^{-}}-(x + 11)\). Substitute \(x = - 6\) into \(-(x + 11)\), we get \(-(-6 + 11)=-5\).

Answer:

\(\lim_{x\to - 6^{+}}f(x)=5\), \(\lim_{x\to - 6^{-}}f(x)=- 5\)