1. **Stating the problem:** We are asked to find the limits of the function $f(x)$ at various points: as $x$ approaches 0, 1 from the left, 1 from the right, exactly at 1, and as $x$ approaches $+\infty$. 2. **Understanding limits:** The limit $\lim_{x \to a} f(x)$ describes the value that $f(x)$ approaches as $x$ gets arbitrarily close to $a$. 3. **Given graph information:** - At $x=0$, there is a dot at $y=3$, so $f(0)=3$. - As $x$ approaches 1 from the left ($x \to 1^-$), the curve goes up to $y=6$ (solid dot at $(1,6)$). - As $x$ approaches 1 from the right ($x \to 1^+$), the curve comes down through $y=4$ (open circle at $(1,4)$), so the right-hand limit is 4. - The function value at $x=1$ is the solid dot, $f(1)=6$. - As $x \to +\infty$, the curve descends approaching zero asymptotically. 4. **Calculating each limit:** - $\lim_{x \to 0} f(x) = 3$ (value at the dot). - $\lim_{x \to 1^-} f(x) = 6$ (approach from left). - $\lim_{x \to 1^+} f(x) = 4$ (approach from right). - $\lim_{x \to 1} f(x)$ exists only if left and right limits are equal; here they differ, so the limit does not exist. - $\lim_{x \to +\infty} f(x) = 0$ (approaches zero asymptotically). 5. **Summary:** $$ \lim_{x \to 0} f(x) = 3 \\ \lim_{x \to 1^-} f(x) = 6 \\ \lim_{x \to 1^+} f(x) = 4 \\ \lim_{x \to 1} f(x) \text{ does not exist} \\ \lim_{x \to +\infty} f(x) = 0 $$ This shows the function has a jump discontinuity at $x=1$ because the left and right limits differ, and the function value at 1 is 6, matching the left limit.