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Algebraically determine the limits. $$ \begin{array}{l} \lim _{x \rightarrow 0.1}[f(x)-g(x)] \text { when } \\ \qquad \lim _{x \rightarrow 0.1} f(x)=6 \text { and } \lim _{x \rightarrow 0.1} g(x)=3 \end{array} $$

Short Answer

Expert verified
The limit is 3.

Step by step solution

01

Write down the given limits

We are given that \( \lim_{x \to 0.1} f(x) = 6 \) and \( \lim_{x \to 0.1} g(x) = 3 \). We need to find \( \lim_{x \to 0.1} [f(x) - g(x)] \).
02

Apply limit properties

We use the property of limits: \[ \lim_{x \to c} [f(x) - g(x)] = \lim_{x \to c} f(x) - \lim_{x \to c} g(x) \] Applying this property for \( x \to 0.1 \), we get: \[ \lim_{x \to 0.1} [f(x) - g(x)] = \lim_{x \to 0.1} f(x) - \lim_{x \to 0.1} g(x) \]
03

Substitute the known limits

Substitute the known values into the formula: \[ \lim_{x \to 0.1} [f(x) - g(x)] = 6 - 3 \]
04

Calculate the result

Perform the subtraction: \[ 6 - 3 = 3 \]. Therefore, the limit is 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Properties
Limits are a fundamental concept in calculus that help us understand the behavior of functions as the input approaches a certain value. One important aspect of limits is understanding how they behave in operations such as addition, subtraction, multiplication, and division. The property used in the exercise is the limit subtraction property, which states:
  • If \( \lim_{x \to c} f(x) = L \) and \( \lim_{x \to c} g(x) = M \), then \( \lim_{x \to c} [f(x) - g(x)] = L - M \).
This property simplifies finding limits of differences between two functions by allowing us to subtract their separate limits. It shows the power of limit properties in easing complex calculations. Understanding and applying these properties can help solve problems efficiently and accurately.
Substitution in Limits
Substitution in limits is a technique often used when we already know the limits of individual function components. In our exercise, we know:
  • \( \lim_{x \to 0.1} f(x) = 6 \)
  • \( \lim_{x \to 0.1} g(x) = 3 \)
With substitution, we place these known values directly into the limit expressions. After identifying which limits belong to which parts of our expression, we substitute them into the equation:
\[ \lim_{x \to 0.1} [f(x) - g(x)] = 6 - 3 \] This is a very efficient way to quickly determine the limits of more complex expressions. It's vital to ensure that the functions are continuous at the point of interest, as substitution is valid only then. In this example, substituting known limits streamlines the problem-solving process, making calculations straightforward and manageable.
Arithmetic Operations with Limits
Arithmetic operations with limits include adding, subtracting, multiplying, and sometimes dividing the limits of functions. This concept makes it easier to discern complex behavior between functions. In the provided exercise, subtraction was key:
  • We use \( \lim_{x \to 0.1} [f(x) - g(x)] \) to find the limit of the difference.
The calculation \[6 - 3 = 3 \] is a direct result of performing arithmetic operations on limits. By applying such operations efficiently, we can solve more complicated limit problems without needing to reinvent calculus. These operations are a vital part of calculus, providing a reliable way to handle multiple functions and understand their collective behavior. Understanding these operations is fundamental in mastering limits and calculus.

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