Math Problem Statement

Solve

Calculate the one-sided limits at x = 2 and determine whether the function is continuous at x = 2. Is the function piecewise continuous in R?

Solution

Let's solve the given problem step by step.

We are given the piecewise function:

\begin{cases} \frac{3x^2 - 4x - 4}{x - 2}, & x < 2 \\ 2\sqrt{x}, & x \geq 2 \end{cases}$$ ### Part (a) - Calculating the one-sided limits at $$x = 2$$ and determining continuity. #### 1. **Right-hand limit** as $$x \to 2^+$$: For $$x \geq 2$$, the function is defined as $$f(x) = 2\sqrt{x}$$. To find the limit from the right at $$x = 2$$: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 2\sqrt{x} = 2\sqrt{2} = 2\sqrt{2}$$ #### 2. **Left-hand limit** as $$x \to 2^-$$: For $$x < 2$$, the function is $$f(x) = \frac{3x^2 - 4x - 4}{x - 2}$$. We need to compute: $$\lim_{x \to 2^-} \frac{3x^2 - 4x - 4}{x - 2}$$ The numerator is $$3x^2 - 4x - 4$$. Let's factorize this polynomial. Using the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-4)}}{2(3)} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}$$ Thus, the roots are $$x = 2$$ and $$x = -\frac{2}{3}$$. So the factorization of the quadratic becomes: $$3x^2 - 4x - 4 = 3(x - 2)\left(x + \frac{2}{3}\right)$$ Now, we can simplify the limit expression: $$\lim_{x \to 2^-} \frac{3(x - 2)\left(x + \frac{2}{3}\right)}{x - 2}$$ Canceling out the $$(x - 2)$$ terms: $$\lim_{x \to 2^-} 3\left(x + \frac{2}{3}\right) = 3\left(2 + \frac{2}{3}\right) = 3 \times \frac{8}{3} = 8$$ #### 3. **Conclusion about continuity** at $$x = 2$$: The right-hand limit is $$2\sqrt{2}$$ and the left-hand limit is $$8$$. Since: $$\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$$ The function is **not continuous** at $$x = 2$$. #### 4. **Is the function piecewise continuous?** Even though the function is not continuous at $$x = 2$$, it is continuous on the intervals $$(-\infty, 2)$$ and $$[2, \infty)$$, meaning it is piecewise continuous on $$\mathbb{R}$$. --- Would you like any details or further clarifications? ### Related Questions: 1. What are the conditions for a function to be continuous at a point? 2. How does factoring polynomials help in simplifying limits? 3. Can a function be discontinuous at a point but still be integrable? 4. What are some examples of piecewise functions that are continuous at their boundary points? 5. How can we graphically represent the one-sided limits and continuity of this function? ### Tip: When finding the limit of a rational function, always check for common factors in the numerator and denominator to simplify the expression before substituting the limit value.

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Math Problem Analysis

Mathematical Concepts

Limits
Piecewise Functions
Continuity

Formulas

Limit as x -> 2+: lim(x→2+) 2√x = 2√2
Limit as x -> 2-: lim(x→2-) (3x^2 - 4x - 4)/(x - 2)

Theorems

Limit definition
Continuity definition

Suitable Grade Level

University level (Calculus I)

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