Question

$$\lim_{x \to -5} f(x) = 4 \quad \lim_{x \to 5} f(x) = 2 \quad \lim_{x \to 5} g(x) = 5$$ the table above gives selected limits of the functions $f$ and $g$. what is $\lim_{x \to 5} \left( f(-x) + 3g(x) \
ight)$? \

$$\begin{tabular}{|c|} \\hline a \\ 19 \\\\ \\hline b \\ 17 \\\\ \\hline c \\ 13 \\\\ \\hline d \\ 9 \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Analyze \(\lim_{x \to 5} f(-x)\)

To find \(\lim_{x \to 5} f(-x)\), we can use the substitution \(t=-x\). When \(x \to 5\), then \(t \to - 5\). So \(\lim_{x \to 5} f(-x)=\lim_{t \to - 5} f(t)\). From the table, we know that \(\lim_{x \to - 5} f(x) = 4\), so \(\lim_{x \to 5} f(-x)=4\)

Step2: Analyze \(\lim_{x \to 5}3g(x)\)

Using the constant multiple rule of limits, \(\lim_{x \to a}(k\cdot h(x))=k\cdot\lim_{x \to a}h(x)\) (where \(k\) is a constant). Here \(k = 3\) and \(h(x)=g(x)\) and \(a = 5\). We know that \(\lim_{x \to 5}g(x)=5\), so \(\lim_{x \to 5}3g(x)=3\times\lim_{x \to 5}g(x)=3\times5 = 15\)

Step3: Analyze \(\lim_{x \to 5}(f(-x)+3g(x))\)

Using the sum rule of limits, \(\lim_{x \to a}(h(x)+k(x))=\lim_{x \to a}h(x)+\lim_{x \to a}k(x)\). Let \(h(x)=f(-x)\) and \(k(x) = 3g(x)\) and \(a = 5\). Then \(\lim_{x \to 5}(f(-x)+3g(x))=\lim_{x \to 5}f(-x)+\lim_{x \to 5}3g(x)\)

Substitute the values we found in step 1 and step 2: \(4 + 15=19\)

Answer:

A. 19