Problem 43 - Calculus for Biology and Medicine

Question

In Problems 29-48, find the limits. $$ \lim _{x \rightarrow-2} \frac{1}{\sqrt{5 x^{2}-4}} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{4} \).

Step 1: Understand the Problem

We need to find the limit \( \lim_{x \to -2} \frac{1}{\sqrt{5x^2 - 4}} \). This involves determining the behavior of the function as \( x \) approaches \( -2 \).

Step 2: Substitute the Value

Substitute \( x = -2 \) directly into the expression \( \sqrt{5x^2 - 4} \) to see if it yields a defined value: \( 5(-2)^2 - 4 = 5(4) - 4 = 20 - 4 = 16 \).

Step 3: Simplify the Limit Expression

The expression becomes \( \frac{1}{\sqrt{16}} \), which simplifies to \( \frac{1}{4} \).

Step 4: Verify the Continuity

Ensure that \( \sqrt{5x^2 - 4} \) is defined and continuous around \( x = -2 \). Since the square root of a positive number is a real number and \( 5x^2 - 4 \) is positive around \( x = -2 \), the expression is continuous at \( x = -2 \).

Step 5: Find the Limit

Since the expression is continuous and we have simplified it to \( \frac{1}{\sqrt{16}} \), the limit as \( x \) approaches \(-2\) is \( \frac{1}{4} \).

Key Concepts

Continuity

Continuity in mathematics refers to the property of a function that allows it to be drawn without lifting the pencil from the paper. A continuous function has no gaps, jumps, or vertical asymptotes. This quality is particularly important when dealing with limits, as it ensures that the behavior of the function near a particular point is predictable.
For a function to be continuous at a point, three conditions must be satisfied:

• The function is defined at the point.
• The limit of the function exists as it approaches that point.
• The limit of the function matches the value of the function at that point.

In the given problem, we are focusing on the function \( \frac{1}{\sqrt{5x^2 - 4}} \) as \( x \) approaches \(-2\). For this function, the component \( \sqrt{5x^2 - 4} \) must be positive to keep the function value real and defined. The calculation of \( \sqrt{5(-2)^2 - 4} \) results in \( \sqrt{16} \), which is a positive number, indicating the function is defined and hence continuous at \( x = -2 \).

Substitution Method

The substitution method is a straightforward way to find the limit of a function as it approaches a certain point. If a function is continuous at the point in question, substitution can be very effective. This method involves directly replacing the variable with the value it is approaching in the given function.
Consider the problem: \( \lim_{x \to -2} \frac{1}{\sqrt{5x^2 - 4}} \). By substituting \( x = -2 \) into the expression, we find it simplifies as follows:

• Calculate \( 5(-2)^2 - 4 = 16 \).
• Compute \( \sqrt{16} \) which equals 4.
• Arrive at \( \frac{1}{4} \) for the limit.

This method assumes that the function behaves well around the point of interest, making it a simple but powerful tool for continuous functions. It helps students grasp the concept of limits more intuitively by directly observing the behavior at the point of substitution.

Square Root Function

The square root function, denoted as \( \sqrt{x} \), is a common mathematical function that outputs a non-negative value, considering only the positive root of any positive input number. The understanding of the square root function is crucial when evaluating expressions involving radicals, especially in limit problems.
Knowing the properties of a square root function can help us solve many mathematical challenges effectively:

• It is only defined for non-negative numbers, as the square root of a negative number is not real.
• It simplifies expressions involving perfect squares, like \( \sqrt{16} = 4 \).
• The function is continuous wherever it is defined in the real number system, such as domains \( x \geq 0 \).

In our example, the expression \( \sqrt{5x^2 - 4} \) needs to be positive since it is under a square root. For \( x = -2\), the calculation results in \( \sqrt{16} \), a real, positive number. This ensures the function is continuous where needed and allows us to successfully find the limit using the substitution method.