Problem 13 - Calculus: Early Transcendentals

Question

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number \( a \). \( p(v) = 2 \sqrt{3v^2 + 1}, \hspace{5mm} a = 1 \)

Step-by-Step Solution

Answer

The function \( p(v) = 2 \sqrt{3v^2 + 1} \) is continuous at \( v = 1 \) as both the function value and the limit are 4.

Step 1: Understand the Definition of Continuity

A function \( f(x) \) is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \). This means the limit of the function as \( x \) approaches \( a \) must equal the function's value at \( a \).

Step 2: Calculate the Value of the Function at a

To find \( f(a) \), substitute \( a = 1 \) into the function \( p(v) = 2 \sqrt{3v^2 + 1} \). \[ p(1) = 2 \sqrt{3(1)^2 + 1} = 2 \sqrt{3 + 1} = 2 \sqrt{4} = 2 \times 2 = 4. \] Hence, \( p(1) = 4 \).

Step 3: Find the Limit of the Function as v approaches a

Next, find \( \lim_{v \to 1} 2 \sqrt{3v^2 + 1} \). Since the function involves a simple polynomial inside a square root, it is continuous for real numbers, and we can directly substitute \( v = 1 \). \[ \lim_{v \to 1} 2 \sqrt{3v^2 + 1} = 2 \sqrt{3(1)^2 + 1} = 2 \sqrt{3 + 1} = 2 \sqrt{4} = 4. \]

Step 4: Verify Continuity Condition

Finally, compare \( \lim_{v \to a} p(v) \) with \( f(a) \). Both are equal to 4. Therefore, since \( \lim_{v \to 1} p(v) = p(1) = 4 \), the function is continuous at \( v = 1 \).

Key Concepts

Definition of Continuity

In mathematics, understanding when a function is continuous is crucial to analyzing its behavior. A function, denoted as \( f(x) \), is said to be continuous at a certain point \( a \) if the limit of the function as \( x \) approaches \( a \) is equal to the function's value at \( a \). This can be formally expressed as \( \lim_{x \to a} f(x) = f(a) \).

• This means that there should be no abrupt jumps or holes at the point \( a \).
• The graph of the function should be smooth and unbroken at that location.
• For continuity, it is also essential that \( f(a) \) is defined and real.

Understanding this definition helps determine whether more complex functions, such as rational or piecewise functions, are continuous at various points.

Properties of Limits

The rules governing limits are fundamental for determining continuity. These properties make handling complex expressions much simpler. Some key properties include:

• The limit of a sum is the sum of the limits: \( \lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \).
• The limit of a product is the product of the limits: \( \lim_{x \to a} (f(x) \times g(x)) = \lim_{x \to a} f(x) \times \lim_{x \to a} g(x) \).
• The limit of a quotient is the quotient of the limits, given the denominator's limit is not zero.

These properties are particularly useful for simplifying expressions before substitution. As in our exercise, they allow us to replace a variable with a number directly, provided it doesn't result in any undefined operations.

Polynomial Functions

Polynomials are a special class of functions often easy to work with due to their smooth nature. A polynomial function can be expressed in terms of its variables raised to non-negative integer powers and has the form \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \).
There are several noteworthy characteristics of polynomial functions:

• They are continuous everywhere on their domain. There are no gaps or jumps in their graphs.
• They have the property that the limit as \( x \) approaches any value \( a \) is simply the function evaluated at \( a \).
• This property, combined with the operations allowed by limit properties, makes them ideal for solving continuity problems.

The simplicity of polynomials, like the expression inside the square root in our example, makes them straightforward when checking for continuity.

Square Roots

The square root function adds an interesting layer to continuity. It takes non-negative values from a given function and returns the number whose square is equal.
Here are some important points about square roots in the context of continuity:

• Square roots are continuous wherever their arguments (the expression inside the root) are non-negative.
• They require careful consideration of domain since the square root of a negative number is not defined in real number space.
• In our example, \( \sqrt{3v^2 + 1} \), the expression within is always positive for any real \( v \), ensuring continuity.

Consequently, in exercises involving square roots, confirming that the under-root expression does not negate simplifies continuity verification substantially. The function \( 2 \sqrt{3v^2 + 1} \) simply translates and scales subsequent results of the polynomial under the root, without affecting continuity.