Math Problem Statement

solve

Find the Limits of the sets presented on the left. Draw the Graph as well.

Solution

The image is displayed. Let's go through the problems one by one to solve the limits and then address the graphs.

9. lim⁡x→0x3X\lim_{x \to 0} \frac{x^3}{X}

To solve this: lim⁡x→0x3x=lim⁡x→0x2=0\lim_{x \to 0} \frac{x^3}{x} = \lim_{x \to 0} x^2 = 0 Thus, the limit is 00.

10. lim⁡x→1x4−1x2−1\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1}

Factor both the numerator and the denominator: x4−1=(x2−1)(x2+1)andx2−1=(x−1)(x+1)x^4 - 1 = (x^2 - 1)(x^2 + 1) \quad \text{and} \quad x^2 - 1 = (x - 1)(x + 1) So, we have: (x2−1)(x2+1)x2−1=x2+1\frac{(x^2 - 1)(x^2 + 1)}{x^2 - 1} = x^2 + 1 Now, taking the limit as x→1x \to 1: lim⁡x→1(x2+1)=12+1=2\lim_{x \to 1} (x^2 + 1) = 1^2 + 1 = 2 Thus, the limit is 22.

11. lim⁡x→0f(x)\lim_{x \to 0} f(x), where f(x)f(x) is piecewise:

x + 3, & x < 0 \\ -x + 3, & x \geq 0 \end{cases}$$ We need to check the left-hand limit ($$x \to 0^-$$) and right-hand limit ($$x \to 0^+$$): - For $$x \to 0^-$$ (from the left), use $$f(x) = x + 3$$: $$\lim_{x \to 0^-} (x + 3) = 0 + 3 = 3$$ - For $$x \to 0^+$$ (from the right), use $$f(x) = -x + 3$$: $$\lim_{x \to 0^+} (-x + 3) = -0 + 3 = 3$$ Since both the left-hand and right-hand limits are equal, the limit is: $$\lim_{x \to 0} f(x) = 3$$ --- ### 12. $$\lim_{x \to 2} f(x)$$, where $$f(x)$$ is piecewise: $$f(x) = \begin{cases} x, & x < 2 \\ x + 1, & x \geq 2 \end{cases}$$ - For $$x \to 2^-$$ (from the left), use $$f(x) = x$$: $$\lim_{x \to 2^-} x = 2$$ - For $$x \to 2^+$$ (from the right), use $$f(x) = x + 1$$: $$\lim_{x \to 2^+} (x + 1) = 2 + 1 = 3$$ Since the left-hand and right-hand limits are different, the limit does not exist. --- ### 13. $$\lim_{x \to 2} f(x)$$, where $$f(x)$$ is piecewise: $$f(x) = \begin{cases} x^2 - 2x, & x < 2 \\ 1, & x = 2 \\ x^2 - 6x + 8, & x > 2 \end{cases}$$ - For $$x \to 2^-$$ (from the left), use $$f(x) = x^2 - 2x$$: $$\lim_{x \to 2^-} (x^2 - 2x) = 2^2 - 2(2) = 4 - 4 = 0$$ - For $$x \to 2^+$$ (from the right), use $$f(x) = x^2 - 6x + 8$$: $$\lim_{x \to 2^+} (x^2 - 6x + 8) = 2^2 - 6(2) + 8 = 4 - 12 + 8 = 0$$ Since both the left-hand and right-hand limits are equal to 0, the limit is: $$\lim_{x \to 2} f(x) = 0$$ --- Would you like detailed explanations on any of these steps or graphs for the functions? Let me know! Here are some questions for further exploration: 1. How would you factor other rational functions when solving limits? 2. Can limits always be evaluated by direct substitution? 3. How can piecewise functions cause a limit to not exist? 4. What is the significance of left-hand and right-hand limits? 5. How can you verify limits graphically? **Tip**: Always check both left and right limits when dealing with piecewise functions, as this helps identify if a limit exists at a point.

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Math Problem Analysis

Mathematical Concepts

Limits
Piecewise Functions
Rational Functions

Formulas

Limit definition: lim_{x -> c} f(x)
Factoring techniques for polynomials
Piecewise function evaluation

Theorems

Limit Laws
Squeeze Theorem
Left-hand and Right-hand Limits

Suitable Grade Level

Grades 11-12

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