Problem 99
Question
In Exercises 99-102, determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that shows it is false. \(\lim _{x \rightarrow 2}\left(\frac{3 x}{x-2}-\frac{2}{x-2}\right)=\lim _{x \rightarrow 2} \frac{3 x}{x-2}-\lim _{x \rightarrow 2} \frac{2}{x-2}\).
Step-by-Step Solution
Verified Answer
The statement is false because the given expression, \(\lim_{x \rightarrow 2} \frac{3x - 2}{x - 2}\), is not equal to the sum of the individual limits, \(\lim _{x \rightarrow 2} \frac{3 x}{x-2} - \lim _{x \rightarrow 2} \frac{2}{x-2}\), as both of these limits are undefined due to the denominator approaching zero.
1Step 1: Simplify the given expression
Given expression: \(\lim _{x \rightarrow 2}\left(\frac{3 x}{x-2}-\frac{2}{x-2}\right)\)
We can write it as: \(\lim_{x \rightarrow 2} \frac{3x - 2}{x - 2}\)
2Step 2: Apply the limit property
According to the property, if a limit exists then \(\lim_{x \rightarrow a} (f(x) + g(x)) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} g(x)\)
Thus, we write: \(\lim _{x \rightarrow 2} \frac{3 x}{x-2} - \lim _{x \rightarrow 2} \frac{2}{x-2}\)
3Step 3: Evaluate the individual limits
Now, we will evaluate the limits individually:
1. \(\lim _{x \rightarrow 2} \frac{3 x}{x-2}\)
As x approaches 2, the denominator approaches zero, hence the limit is undefined.
2. \(\lim _{x \rightarrow 2} \frac{2}{x-2}\)
Similar to the previous limit, as x approaches 2, the denominator approaches zero, hence this limit is also undefined.
4Step 4: Compare both results
Comparing the results, we have:
\(\lim _{x \rightarrow 2}\left(\frac{3 x}{x-2}-\frac{2}{x-2}\right)\) in simplified form from step 1 is \(\lim_{x \rightarrow 2} \frac{3x - 2}{x - 2}\),
which is not equal to \(\lim _{x \rightarrow 2} \frac{3 x}{x-2}-\lim _{x \rightarrow 2} \frac{2}{x-2}\) which is undefined for both limits.
Hence, the statement is
5Step 5: False
The statement is false as the limits cannot be found individually due to the denominator approaching zero.
Key Concepts
Limit UndefinedLimit PropertiesEvaluating LimitsSimplifying Expressions
Limit Undefined
In calculus, when evaluating limits, you might come across situations where the limit is undefined. This typically happens when the expression involves a division by zero. As x approaches a certain value, if the denominator of a fraction approaches zero, the value of the function can become infinitely large or behave unpredictably. This is referred to as the limit being undefined.
In such cases, it is crucial to analyze the behavior of the numerator and the denominator precisely. Both limits \(\lim _{x \rightarrow 2} \frac{3 x}{x-2}\) and \(\lim _{x \rightarrow 2} \frac{2}{x-2}\) involve a denominator that becomes zero as \( x \) approaches 2, leading to an undefined limit.
In such cases, it is crucial to analyze the behavior of the numerator and the denominator precisely. Both limits \(\lim _{x \rightarrow 2} \frac{3 x}{x-2}\) and \(\lim _{x \rightarrow 2} \frac{2}{x-2}\) involve a denominator that becomes zero as \( x \) approaches 2, leading to an undefined limit.
Limit Properties
The properties of limits are fundamental in simplifying and evaluating complex limit expressions. One crucial property is that the limit of a sum (or difference) of functions can be split into the sum (or difference) of their individual limits, provided those limits exist. This property can be expressed as: \(\lim_{x \rightarrow a} (f(x) + g(x)) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} g(x)\).
This is powerful as it allows us to break down complex expressions into manageable parts. However, if any part has an undefined limit, the entire expression's limit may also be undefined or require more advanced techniques to resolve.
This is powerful as it allows us to break down complex expressions into manageable parts. However, if any part has an undefined limit, the entire expression's limit may also be undefined or require more advanced techniques to resolve.
Evaluating Limits
Evaluating limits is a fundamental skill in calculus. It involves determining what value a function approaches as the input nears a specific point. Direct substitution is often the first step in evaluating limits. However, when direct substitution leads to an indeterminate form like \(\frac{0}{0}\), other techniques such as simplification or L'Hôpital's Rule might be necessary.
In the exercise given, trying to find the limits directly led to an undefined value due to division by zero, highlighting the importance of recognizing such situations and employing strategies such as algebraic manipulation or considering the behavior of the function around the point in question.
In the exercise given, trying to find the limits directly led to an undefined value due to division by zero, highlighting the importance of recognizing such situations and employing strategies such as algebraic manipulation or considering the behavior of the function around the point in question.
Simplifying Expressions
Simplifying expressions is a critical step in resolving limits. By manipulating the algebraic form, you might remove obstacles like division by zero that lead to undefined limits. This can involve factoring, cancelling terms, or rationalizing in order to transform the expression into a form that allows easier limit calculation.
In the original exercise, simplifying \(\lim_{x \rightarrow 2} \left(\frac{3x}{x-2} - \frac{2}{x-2}\right)\) into a single fraction \(\frac{3x - 2}{x - 2}\) was a key step. Although the denominator still approaches zero, simplifying helps reveal the function's behavior more clearly, though in this case, both forms lead to an undefined limit as \( x \) approaches 2.
In the original exercise, simplifying \(\lim_{x \rightarrow 2} \left(\frac{3x}{x-2} - \frac{2}{x-2}\right)\) into a single fraction \(\frac{3x - 2}{x - 2}\) was a key step. Although the denominator still approaches zero, simplifying helps reveal the function's behavior more clearly, though in this case, both forms lead to an undefined limit as \( x \) approaches 2.
Other exercises in this chapter
Problem 95
Suppose that \(f(x)
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Show that \(\lim _{x \rightarrow a} \cos x=\cos a\). (See the hint for Exercise 97.)
View solution Problem 100
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give \(a n\) example that shows it is false. If \(f
View solution Problem 101
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that shows it is false. If \(\lim _
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