Math Problem Statement

Math 1341 Calculus 1 Homework Chapter 1.3: Limit problems including factoring, piecewise functions, and limits at infinity.

Solution

Let's solve the calculus problems one by one from the image you've uploaded:

Problem 1:

limx1x2+3x+2x+1\lim_{x \to -1} \frac{x^2 + 3x + 2}{x+1} We can first simplify the given expression by factoring the numerator: x2+3x+2=(x+1)(x+2)x^2 + 3x + 2 = (x + 1)(x + 2) Thus, the expression becomes: (x+1)(x+2)x+1\frac{(x + 1)(x + 2)}{x + 1} For x1x \neq -1, we can cancel the x+1x + 1 terms: (x+1)(x+2)x+1=x+2\frac{(x + 1)(x + 2)}{x + 1} = x + 2 Now, we can directly substitute x=1x = -1 into the simplified expression: x+2=1+2=1x + 2 = -1 + 2 = 1 Therefore, limx1x2+3x+2x+1=1\lim_{x \to -1} \frac{x^2 + 3x + 2}{x+1} = 1


Problem 2:

3x + 1 & \text{if } x \leq 0 \\ 2 & \text{if } x > 0 \end{cases}$$ We need to check the left-hand limit and the right-hand limit separately: - For the **left-hand limit** $$x \to 0^{-}$$, we use $$f(x) = 3x + 1$$: $$\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (3x + 1) = 1$$ - For the **right-hand limit** $$x \to 0^{+}$$, we use $$f(x) = 2$$: $$\lim_{x \to 0^{+}} f(x) = 2$$ Since the left-hand limit and right-hand limit are not equal, the limit **does not exist**: $$\lim_{x \to 0} f(x) \text{ does not exist.}$$ --- ### Problem 3: $$\lim_{x \to -1} \frac{x^2}{\sqrt{x+2}}$$ First, substitute $$x = -1$$ directly: $$\frac{(-1)^2}{\sqrt{-1 + 2}} = \frac{1}{\sqrt{1}} = 1$$ Thus, the limit is: $$\lim_{x \to -1} \frac{x^2}{\sqrt{x+2}} = 1$$ --- ### Problem 4 (not shown): We will skip this problem as it is not visible in the uploaded image. --- ### Problem 5: $$\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}}$$ We can divide both the numerator and the denominator by $$x$$: $$\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}} = \lim_{x \to \infty} \frac{x(3 - \frac{2}{x})}{x\sqrt{1 - \frac{10}{x^2}}}$$ This simplifies to: $$\lim_{x \to \infty} \frac{3 - \frac{2}{x}}{\sqrt{1 - \frac{10}{x^2}}}$$ As $$x \to \infty$$, $$\frac{2}{x} \to 0$$ and $$\frac{10}{x^2} \to 0$$, so the expression becomes: $$\frac{3 - 0}{\sqrt{1 - 0}} = \frac{3}{1} = 3$$ Thus, the limit is: $$\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}} = 3$$ --- ### Summary: 1. $$\lim_{x \to -1} \frac{x^2 + 3x + 2}{x+1} = 1$$ 2. $$\lim_{x \to 0} f(x)$$ does not exist. 3. $$\lim_{x \to -1} \frac{x^2}{\sqrt{x+2}} = 1$$ 5. $$\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}} = 3$$ Do you have any questions, or would you like more details? ### Related Questions: 1. What is the general process for solving limits using factoring? 2. How do piecewise functions affect the existence of limits? 3. What are the strategies for evaluating limits as $$x \to \infty$$? 4. How can you recognize when a limit does not exist from a graph? 5. How does L'Hôpital's rule apply to evaluating difficult limits? ### Tip: Always simplify the expression first, if possible, before trying to directly evaluate a limit. This often helps avoid indeterminate forms.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Factoring
Piecewise Functions
Limits at Infinity
Square Roots

Formulas

lim_{x \to a} \frac{f(x)}{g(x)}
lim_{x \to \infty} \frac{f(x)}{g(x)}
Factoring Quadratics: x^2 + bx + c = (x + p)(x + q)

Theorems

Limit Definition
Limit of Piecewise Functions
Limit Laws

Suitable Grade Level

College (Calculus 1)