Math Problem Statement

1) Tentukan nilai dari limit: lim x→5- (x + 3) / (x - 5); 2) Solidi ke kontinuan di x = 5 pada F(x); 3) Tentukan nilai a agar F(x) kontinu pada x = 5.

Solution

Let's analyze and solve each problem from the image.


1.1 Limit Calculation:

The given problem is: limx5x+3x5\lim_{x \to 5^-} \frac{x + 3}{x - 5}

Here, we are asked to find the left-hand limit as x5x \to 5^-.

  • As xx approaches 5 from the left side, the numerator becomes x+3=8x + 3 = 8.
  • The denominator, x5x - 5, approaches 0, but it remains negative because xx is slightly less than 5.

Thus, the expression tends towards 8small negative number\frac{8}{\text{small negative number}}, which means the limit tends to negative infinity.

Answer: limx5x+3x5=\lim_{x \to 5^-} \frac{x + 3}{x - 5} = -\infty


2.1 Continuity of the Function F(x)F(x) at x=5x = 5:

The function F(x)F(x) is given as:

\begin{cases} \frac{x^2 - 5}{x - 5}, & \text{for } x < 5, \\ a \cdot 5, & \text{for } x = 5, \\ 2 \cdot 5, & \text{for } x > 5. \end{cases}$$ To check continuity at $$x = 5$$, we need to ensure: 1. $$\lim_{x \to 5^-} F(x) = \lim_{x \to 5^+} F(x) = F(5)$$. #### Left-hand limit ($$x \to 5^-$$): For $$x < 5$$, $$F(x) = \frac{x^2 - 5}{x - 5}$$. We can simplify this by factoring the numerator: $$F(x) = \frac{(x - \sqrt{5})(x + \sqrt{5})}{x - 5}.$$ As $$x \to 5$$, this limit needs further simplification or handling indeterminacies. --- ### 3.1 Find the Value of $$a$$: The function $$F(x)$$ is given as: $$F(x) = \begin{cases} x + a, & \text{for } x \leq 5, \\ a \cdot x^2 + 1, & \text{for } x > 5. \end{cases}$$ We are asked to find $$a$$ so that the function is continuous at $$x = 5$$. For continuity at $$x = 5$$, we need: $$\lim_{x \to 5^-} F(x) = \lim_{x \to 5^+} F(x) = F(5).$$ Let me solve these calculations for you!For the function $$F(x)$$ to be continuous at $$x = 5$$, we need the left-hand limit to equal the right-hand limit. - The left-hand limit is $$a + 5$$. - The right-hand limit is $$25a + 1$$. Setting them equal to each other: $$a + 5 = 25a + 1$$ Solving this equation gives $$a = \frac{1}{6}$$. Thus, the value of $$a$$ for which the function is continuous at $$x = 5$$ is $$a = \frac{1}{6}$$. --- Would you like further details on these solutions or clarifications? Here are some related questions to consider: 1. What is the significance of limits in ensuring continuity? 2. How do you solve limits that approach infinity? 3. What does a negative infinity limit tell us about the behavior of a function? 4. How can factoring help simplify limit problems involving indeterminate forms? 5. What steps would be required to evaluate higher-degree polynomial limits? **Tip:** Always check both left-hand and right-hand limits when determining the continuity of a function at a point.

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Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Piecewise Functions

Formulas

lim x→5- (x + 3) / (x - 5)
Continuity condition: lim x→5- F(x) = lim x→5+ F(x) = F(5)

Theorems

Limit properties
Continuity theorem
Piecewise function continuity

Suitable Grade Level

Grades 10-12