Problem 11 - Contemporary Precalculus

Question

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number. $$f(x)=\frac{x \sqrt{x}}{(x-6)^{2}} \quad \text { at } x=36$$

Step-by-Step Solution

Answer

Answer: Yes, the function is continuous at x = 36 because the function value, left limit, and right limit are all equal to $$\frac{12}{50}$$ at x = 36.

Step 1: Identify the function and the given number

We are given the function: $$ f(x)=\frac{x \sqrt{x}}{(x-6)^{2}} $$ And we are asked to show continuity at x = 36.

Step 2: Evaluate the function at x = 36

We first find the value of the function at x = 36: $$ f(36)=\frac{36 \sqrt{36}}{(36-6)^{2}} = \frac{36 \cdot 6}{(30)^{2}} =\frac{216}{900} = \frac{12}{50} $$

Step 3: Evaluate the left limit at x = 36

The left limit is the limit as x approaches 36 from the left. We evaluate this limit using the properties of limits: $$ \lim_{x \to 36^{-}} \frac{x \sqrt{x}}{(x-6)^{2}} $$ As x approaches 36, the function has no singularities, and the numerator and denominator are both well-defined. So, we can replace x with 36 directly to find the left limit: $$ \lim_{x \to 36^{-}} \frac{x \sqrt{x}}{(x-6)^{2}} = \frac{36 \sqrt{36}}{(36-6)^{2}} = \frac{12}{50} $$

Step 4: Evaluate the right limit at x = 36

The right limit is the limit as x approaches 36 from the right. We evaluate this limit using the properties of limits: $$ \lim_{x \to 36^{+}} \frac{x \sqrt{x}}{(x-6)^{2}} $$ As x approaches 36, the function has no singularities, and the numerator and denominator are both well-defined. So, we can replace x with 36 directly to find the right limit: $$ \lim_{x \to 36^{+}} \frac{x \sqrt{x}}{(x-6)^{2}} = \frac{36 \sqrt{36}}{(36-6)^{2}} = \frac{12}{50} $$

Step 5: Compare function value, left limit, and right limit

Now, we compare the function value, left limit, and right limit at x = 36: Function value at x = 36: $$\frac{12}{50}$$ Left limit at x = 36: $$\frac{12}{50}$$ Right limit at x = 36: $$\frac{12}{50}$$ Since the function value, left limit, and right limit are all equal, the function is continuous at x = 36.

Key Concepts

Properties of Limits

Understanding the properties of limits is essential when analyzing the behavior of a function as it approaches a specific point.

A limit describes the value that a function approaches as the input (or x-value) approaches a certain value. Crucial properties of limits used in calculus and precalculus include the following:

• Limit of a Constant: The limit of a constant is the constant itself, ewline ewline \(\text{lim}_{x \to a} c = c\), where c is a constant and a is the point we are approaching.
• Limit of a Linear Function: For a linear function, the limit as x approaches any number is just the value of the function at that point, ewline ewline \(\text{lim}_{x \to a} (mx+b) = ma + b\), where m is the slope and b is the y-intercept.
• Sum and Difference: The limit of a sum or difference is the sum or difference of the limits, ewline ewline \(\text{lim}_{x \to a} (f(x) \text{pm} g(x)) = \text{lim}_{x \to a} f(x) \text{pm} \text{lim}_{x \to a} g(x)\).
• Product: The limit of a product is the product of the limits, ewline ewline \(\text{lim}_{x \to a} [f(x)g(x)] = (\text{lim}_{x \to a} f(x))(\text{lim}_{x \to a} g(x))\).
• Quotient: The limit of a quotient is the quotient of the limits (provided the limit of the denominator is not zero), ewline ewline \(\text{lim}_{x \to a} \frac{f(x)}{g(x)} = \frac{\text{lim}_{x \to a} f(x)}{\text{lim}_{x \to a} g(x)}\), with the condition that \(\text{lim}_{x \to a} g(x) neq 0\).

Using these properties allows us to evaluate limits more easily without computing the limit from first principles every single time. For instance, in the provided exercise, the function ewline ewline \(\frac{x \text{sqrt}{x}}{(x-6)^{2}}\) is a combination of polynomial and radical functions for which limits can be evaluated directly by substitution when x approaches the point of interest, in this case, 36.

Continuous Functions

A function is continuous at a point if it satisfies three conditions:

• The function is defined at that point.
• The limit exists as x approaches that point.
• The limit's value is equal to the function's value at that point.

Intuitively, you can think of a function as continuous if you can draw it without lifting your pencil from the paper. Continuity is an important concept because it guarantees that functions behave predictably around the point of continuity and are free from any abrupt changes or 'jumps.'

Regarding the original exercise, the function ewline ewline \(\frac{x \text{sqrt}{x}}{(x-6)^{2}}\), is continuous at \(x = 36\) because it meets all three conditions for continuity at that point. It's defined since both the numerator and the denominator are non-zero; the limits as x approaches 36 from both the left and the right exist and are equal, and both are equal to the function's value at \(x = 36\). This smooth behavior around \(x = 36\) is typical of continuous functions, and understanding this property helps students in precalculus anticipate how functions will behave.

Evaluating Limits

Evaluating limits is a foundational skill in calculus that helps predict where functions are headed, as opposed to where they are or have been. To evaluate a limit, we consider what value the function is approaching, not necessarily what value it reaches. Here are some strategies to evaluate limits:

• Direct Substitution: If a function is continuous at the point of interest, we can evaluate the limit by simply substituting the point into the function.
• Factoring: Factor the expression and simplify to remove common factors before substituting the point of interest.
• Rationalizing: For functions involving square roots, multiply by a conjugate to rationalize the numerator or denominator.
• L'Hôpital's Rule: When a limit presents an indeterminate form like \(0/0\) or \(\text{inf}/\text{inf}\), applying L'Hôpital's Rule by taking derivatives can provide a way forward.

In our exercise, we evaluated the limit of \(\frac{x \text{sqrt}{x}}{(x-6)^{2}}\) as \(x\) approaches 36 by direct substitution, since there were no singularities or indeterminate forms to be addressed. The ease of evaluation by direct substitution is a testament to the continuity of the function at the approached point. Learning to evaluate limits confidently and accurately is an indispensable part of mastering precalculus and calculus.