1. **Problem Statement:** We have a piecewise function $ j(x) $ defined as: $$ j(x) = \begin{cases} f(x) & \text{if } -5 \leq x < 0 \\ g(x) & \text{if } 0 \leq x \leq 1 \\ h(x) = x \ln(x) + k & \text{if } x > 1 \end{cases} $$ We need to analyze continuity at $ x = -4, -2, 0 $, find $ k $ for continuity at $ x=1 $, find horizontal asymptotes of $ f(x) $, and find $ \lim_{x \to \infty} j(x) $. --- 2. **(a) Continuity at $ x = a $:** Recall the definition of continuity at $ x = a $: $$ \lim_{x \to a} j(x) = j(a) $$ This means the left-hand limit (LHL), right-hand limit (RHL), and function value at $ a $ must all be equal. - **At $ a = -4 $:** - Since $ -5 \leq x < 0 $, $ j(x) = f(x) $ on both sides near $ -4 $. - The graph shows a filled point at $ (-4, 2) $ and an open circle at $ (-4, 0) $ on the y-axis. - The function value $ j(-4) = f(-4) = 2 $ (filled point). - The limit from both sides is $ 2 $ because the function is defined by $ f(x) $ on both sides. - Therefore, $ j(x) $ is continuous at $ x = -4 $. - **At $ a = -2 $:** - $ j(x) = f(x) $ for $ x < 0 $, so near $ -2 $, $ j(x) = f(x) $. - The graph shows no discontinuity or open circles near $ -2 $. - Hence, $ j(x) $ is continuous at $ x = -2 $. - **At $ a = 0 $:** - Left side limit: $ \lim_{x \to 0^-} j(x) = \lim_{x \to 0^-} f(x) $. - Right side limit: $ \lim_{x \to 0^+} j(x) = \lim_{x \to 0^+} g(x) $. - From the graph, $ f(0) $ has an open circle at $ y=2 $, so $ f(0) $ is not defined or not equal to 2. - $ g(0) = 2 $ (filled point). - Since $ \lim_{x \to 0^-} f(x) \neq g(0) $, $ j(x) $ is not continuous at $ x=0 $. --- 3. **(b) Continuity at $ x = 1 $:** We want: $$ \lim_{x \to 1^-} j(x) = j(1) = \lim_{x \to 1^+} j(x) $$ - Left limit: $ \lim_{x \to 1^-} j(x) = \lim_{x \to 1^-} g(x) = g(1) = 1 $ (from table). - Right limit: $ \lim_{x \to 1^+} j(x) = \lim_{x \to 1^+} h(x) = \lim_{x \to 1^+} (x \ln x + k) = 1 \cdot \ln 1 + k = 0 + k = k $. - Function value at 1 is $ j(1) = g(1) = 1 $. Set left and right limits equal: $$ 1 = k $$ So, $ k = 1 $. --- 4. **(c) Horizontal asymptotes of $ f(x) $:** From the graph description: - As $ x \to -5 $, $ f(x) $ approaches $ y = 0 $ (open circle at approx 0). - As $ x \to 0^- $, $ f(x) $ approaches $ y = 2 $ (open circle at 2). Therefore, horizontal asymptotes are: $$ y = 0 \quad \text{and} \quad y = 2 $$ --- 5. **(d) $ \lim_{x \to \infty} j(x) $:** For $ x > 1 $, $ j(x) = h(x) = x \ln x + k $. As $ x \to \infty $, $ \ln x \to \infty $ but grows slower than $ x $, so: $$ \lim_{x \to \infty} x \ln x = \infty $$ Thus, $$ \lim_{x \to \infty} j(x) = \infty $$ Interpretation: The function $ j(x) $ grows without bound as $ x $ becomes very large. --- **Final answers:** - (a) Continuous at $ x = -4 $ and $ x = -2 $, not continuous at $ x = 0 $. - (b) $ k = 1 $ for continuity at $ x = 1 $. - (c) Horizontal asymptotes of $ f(x) $ are $ y = 0 $ and $ y = 2 $. - (d) $ \lim_{x \to \infty} j(x) = \infty $.