1. **Problem 1: Determine the truth of limit statements for the function $y=f(x)$ given the graph.** - The graph is piecewise linear with points: $(-1,-1)$ solid, $(0,0)$ open, $(1,1)$ open, $(2,1)$ solid. - Lines connect $(-1,-1)$ to $(0,0)$, $(0,0)$ to $(1,1)$, and $(1,1)$ to $(2,1)$. **Step 1:** Analyze $\lim_{x \to 0} f(x)$. - From the left, as $x \to 0^-$, $f(x) \to 0$. - From the right, as $x \to 0^+$, $f(x) \to 0$. - Both sides approach 0, so $\lim_{x \to 0} f(x) = 0$. **Step 2:** Analyze $\lim_{x \to 1} f(x)$. - From the left, as $x \to 1^-$, $f(x) \to 1$. - From the right, as $x \to 1^+$, $f(x) \to 1$. - Both sides approach 1, so $\lim_{x \to 1} f(x) = 1$. **Step 3:** Check if $\lim_{x \to x_0} f(x)$ exists for every $x_0 \in (-1,1)$. - The function is linear and continuous between points except at $x=0$ and $x=1$ where there are open circles. - However, limits depend on approaching values, not function value at the point. - Since the graph is continuous (no jumps) in $(-1,1)$, limits exist at every $x_0$ in $(-1,1)$. **Step 4:** Evaluate each statement: - a. $\lim_{x \to 0} f(x)$ exists: **True**. - b. $\lim_{x \to 0} f(x) = 0$: **True**. - c. $\lim_{x \to 0} f(x) = 1$: **False**. - d. $\lim_{x \to 1} f(x) = 1$: **True**. - e. $\lim_{x \to 1} f(x) = 0$: **False**. - f. $\lim_{x \to x_0} f(x)$ exists for every $x_0 \in (-1,1)$: **True**. 2. **Problem 2: Explain why $\lim_{x \to 0} \frac{x}{|x|}$ does not exist.** **Step 1:** Consider the left-hand limit as $x \to 0^-$: - For $x<0$, $|x| = -x$, so $$\frac{x}{|x|} = \frac{x}{-x} = -1.$$ - Thus, $\lim_{x \to 0^-} \frac{x}{|x|} = -1$. **Step 2:** Consider the right-hand limit as $x \to 0^+$: - For $x>0$, $|x| = x$, so $$\frac{x}{|x|} = \frac{x}{x} = 1.$$ - Thus, $\lim_{x \to 0^+} \frac{x}{|x|} = 1$. **Step 3:** Since the left-hand and right-hand limits are not equal, $$\lim_{x \to 0} \frac{x}{|x|}$$ **does not exist**. **Final answers:** - Q1: a) True, b) True, c) False, d) True, e) False, f) True. - Q2: The limit does not exist because the left and right limits are different: $-1$ and $1$ respectively.