Problem 104
Question
In Exercises \(101-104,\) use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow-\infty} \frac{1}{x-2}=0 $$
Step-by-Step Solution
Verified Answer
Based on the steps followed, it can be concluded that \(\lim_{x \rightarrow -\infty} \frac{1}{x-2} = 0\) is proven by using the definition of limits at infinity.
1Step 1: Define the limit
The task is to prove that for all \(\varepsilon > 0\), there exists a number \(M\) such that if \(x < M\), then \(|f(x) - L| < \varepsilon\), where \(f(x) = \frac{1}{x-2}\), and \(L = 0\). In this context, \(f(x)\) is \( \frac{1}{x-2}\), \(L\) is 0, \(\varepsilon\) is any positive number, and \(M\) will be determined.
2Step 2: Apply the definition
Using the definition, we must show that for any positive number \(\varepsilon\), we can find a sufficiently negative value \(M\), such that if \(x < M\), then \(|\frac{1}{x-2}| < \varepsilon\). We express the absolute value as \(\frac{-1}{x-2}\) when \(x < 2\).
3Step 3: Find the value of M
We solve the inequality \(\frac{-1}{x-2} < \varepsilon\) for \(x\). Manipulating the inequality, we get \(x < \frac{-1}{\varepsilon} + 2 = M\). Hence, for \(x < M\), the absolute inequality holds.
4Step 4: Draw the conclusion
Since for any given \(\varepsilon > 0\), a corresponding \(M\) can be found that fulfills the requirements of the definition of the limit, it is proven that \(\lim_{x \rightarrow -\infty} \frac{1}{x-2} = 0\).
Key Concepts
Definition of LimitsLimit ProofsCalculus Limits
Definition of Limits
Limits in calculus are a fundamental concept used to define values that functions approach as the input becomes infinitely large or infinitely small. This is crucial when considering behavior at infinity.
A limit at infinity specifically involves studying how functions behave as the input variable grows very large (positively or negatively). For instance, in the exercise given, we explore how the function \( f(x) = \frac{1}{x - 2} \) behaves as \( x \rightarrow -\infty \).
The formal definition of a limit at infinity is as follows: \( \lim_{x \to c} f(x) = L \) means that for every \( \varepsilon > 0 \), there exists a number \( M \) such that if \( x < M \), then the function \( |f(x) - L| < \varepsilon \). This implies the function becomes arbitrarily close to \( L \) as \( x \) increases or decreases without bound.
A limit at infinity specifically involves studying how functions behave as the input variable grows very large (positively or negatively). For instance, in the exercise given, we explore how the function \( f(x) = \frac{1}{x - 2} \) behaves as \( x \rightarrow -\infty \).
The formal definition of a limit at infinity is as follows: \( \lim_{x \to c} f(x) = L \) means that for every \( \varepsilon > 0 \), there exists a number \( M \) such that if \( x < M \), then the function \( |f(x) - L| < \varepsilon \). This implies the function becomes arbitrarily close to \( L \) as \( x \) increases or decreases without bound.
- This definition helps us formalize our intuition about the behavior of functions at the extremes of their domains.
- It is a crucial tool for proving limits rigorously.
Limit Proofs
Epsilon-delta proofs, often referred to as \( \varepsilon - \delta \) proofs, provide a rigorous foundation to establish limits. In context, epsilon (\( \varepsilon \)) represents any small positive number, whereas \( M \) pertains to a sufficiently large or small number that guides us toward proving the limit at infinity.
In our exercise, we manage \( \varepsilon \) and \( M \) to show that \( \lim_{x \to -\infty} \frac{1}{x-2} = 0 \). We say that for each choice of \( \varepsilon > 0 \), there exists an \( M \) such that if \( x < M \), \( |\frac{1}{x-2}| < \varepsilon \).
Here's a simplified view of why this works:
This structured approach confirms the behavior of \( f(x) \) as \( x \rightarrow -\infty \).
In our exercise, we manage \( \varepsilon \) and \( M \) to show that \( \lim_{x \to -\infty} \frac{1}{x-2} = 0 \). We say that for each choice of \( \varepsilon > 0 \), there exists an \( M \) such that if \( x < M \), \( |\frac{1}{x-2}| < \varepsilon \).
Here's a simplified view of why this works:
- Pick \( \varepsilon \), an "acceptable level" for how close \( f(x) \) should be to 0.
- We solve the inequality \( \frac{-1}{x-2} < \varepsilon \) to find an \( M \).
- This translates to \( x < \frac{-1}{\varepsilon} + 2 \), providing us with the required \( M \).
- Therefore, \( f(x) \) gets within an epsilon neighborhood of 0, illustrating that the formal limit holds.
This structured approach confirms the behavior of \( f(x) \) as \( x \rightarrow -\infty \).
Calculus Limits
Calculus limits serve as the backbone for understanding many concepts in calculus, including the behavior of sequences and functions. They set the stage for more complex topics, such as derivatives and integrals.
In calculus, limits allow us to:
Returning to the exercise, the computation of \( \lim_{x \to -\infty} \frac{1}{x-2} = 0 \) illustrates these principles. The limit at \(-\infty\) helps us see that \( f(x) \) "levels out" at 0 as \( x \) decreases indefinitely. This understanding is invaluable when progressing to other calculus concepts. Practical applications abound in physics, engineering, and other sciences where modeling behavior at extreme values is necessary.
In calculus, limits allow us to:
- Analyze the behavior of functions as they approach specific points or infinity.
- Enable us to talk about continuity by describing where functions do not "break".
- Help compute the instantaneous rate of change, which leads to the derivative.
- Aid in understanding the area under curves, which leads to the integral.
Returning to the exercise, the computation of \( \lim_{x \to -\infty} \frac{1}{x-2} = 0 \) illustrates these principles. The limit at \(-\infty\) helps us see that \( f(x) \) "levels out" at 0 as \( x \) decreases indefinitely. This understanding is invaluable when progressing to other calculus concepts. Practical applications abound in physics, engineering, and other sciences where modeling behavior at extreme values is necessary.
Other exercises in this chapter
Problem 102
In Exercises \(101-104,\) use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow \infty} \frac{2}{\sqrt{x}}=0 $$
View solution Problem 103
In Exercises \(101-104,\) use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow-\infty} \frac{1}{x^{3}}=0 $$
View solution Problem 105
Let \(x>0\) and \(n>1\) be real numbers. Prove that \((1+x)^{n}>1+n x\).
View solution Problem 105
Prove that if \(p(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}\) and \(q(x)=b_{m} x^{m}+\cdots+b_{1} x+b_{0}\left(a_{n} \neq 0, b_{m} \neq 0\right),\) then \(\lim _{x \r
View solution