Problem 60
Question
\(60-61 .\) Definition of an infinite limit We write \(\lim _{x \rightarrow a} f(x)=-\infty\) if for any negative number \(M,\) there exists \(\delta >0\) such that $$ f(x) < M \text { whenever } 0 <| x - a |< \delta $$ Use this definition to prove the following statements. $$\lim _{x \rightarrow 1} \frac{-2}{(x-1)^{2}}=-\infty$$
Step-by-Step Solution
Verified Answer
Question: Prove that \(\lim_{x \rightarrow 1} \frac{-2}{(x-1)^{2}} = -\infty\).
Answer: We found the relationship between \(M\) and \(\delta\) such that \(\delta < \sqrt{\frac{-2}{M}}\). This shows that for any negative number \(M\), there exists \(\delta > 0\) such that \(f(x) = \frac{-2}{(x-1)^2} < M\) whenever \(0 < |x-1| < \delta\). Therefore, \(\lim_{x \rightarrow 1} \frac{-2}{(x-1)^{2}} = -\infty\).
1Step 1: Rewrite \(f(x)\) in terms of \(M\) and \(\delta\)
We want to find the relationship between \(M\) and \(\delta\) for the given function \(f(x) = \frac{-2}{(x-1)^2}\), so that \(f(x) < M\) whenever \(0 < |x - 1| < \delta\). We can rewrite the inequality \(f(x) < M\) as:
$$
\frac{-2}{(x-1)^2} < M
$$
2Step 2: Rearrange the inequality to isolate \((x-1)^2\)
Now, in order to get a relationship between \(M\) and \(\delta\), we need to isolate \((x-1)^2\) since that involves our variable \(x\). This will allow us to relate \(M\) to \(0<|x-1|<\delta\).
$$
\frac{-2}{M}<(x-1)^2
$$
3Step 3: Relate \(\delta\) to \(M\)
We know that \(0 < |x-1| < \delta\). Thus, we need to relate \((x-1)^2\) and \(\delta\). We can do this by squaring the right side of the inequality, i.e., \(|x-1|^2<\delta^2\). Thus, \((x-1)^2<\delta^2\). Now, we need to find a \(\delta\) such that \((x-1)^2<\delta^2<\frac{-2}{M}\).
Since \(\delta\) is positive, we can take the square root on both sides of the inequality:
$$
\delta<\sqrt{\frac{-2}{M}}
$$
4Step 4: Use the relationship to verify the definition of an infinite limit
Now, we have found the relationship between \(M\) and \(\delta\). Using this relationship, we can verify the definition of an infinite limit \(\lim_{x \rightarrow 1} \frac{-2}{(x-1)^{2}}=-\infty\).
Given the inequality \(\delta < \sqrt{\frac{-2}{M}}\), we have \(\delta\) defined for any negative number \(M\) such that \(f(x) = \frac{-2}{(x-1)^2} < M\) whenever \(0 < |x-1| < \delta\). This verifies that \(\lim_{x \rightarrow 1} \frac{-2}{(x-1)^{2}} = -\infty\).
Key Concepts
Limit DefinitionAsymptotic BehaviorCalculus Problem SolvingLimit Laws and Techniques
Limit Definition
Understanding the definition of a limit in calculus is fundamental to tackling many types of problems. Essentially, when we write something like \(\lim_{x \rightarrow a} f(x) = L\), we're saying that as \(x\) gets closer and closer to the value \(a\), the function \(f(x)\) gets closer to the limit \(L\). However, there is a special condition when we talk about infinite limits. With an infinite limit, as \(x\) approaches a specific value - let's say \(a\), the function value does not settle to a particular number but rather increases or decreases without bound.
For instance, when we say \(\lim_{x \rightarrow a} f(x) = -\infty\), it means no matter how close \(x\) gets to \(a\), the value of \(f(x)\) will always be less than any given negative number \(M\), provided \(x\) is within a distance \(\delta\) from \(a\). Conveying this concept in simple terms: as \(x\) inches towards \(a\), \(f(x)\) falls off the edge into a pit of infinitely decreasing values, never hitting the bottom.
For instance, when we say \(\lim_{x \rightarrow a} f(x) = -\infty\), it means no matter how close \(x\) gets to \(a\), the value of \(f(x)\) will always be less than any given negative number \(M\), provided \(x\) is within a distance \(\delta\) from \(a\). Conveying this concept in simple terms: as \(x\) inches towards \(a\), \(f(x)\) falls off the edge into a pit of infinitely decreasing values, never hitting the bottom.
Asymptotic Behavior
The notion of asymptotic behavior is like the calculus equivalent of a border that keeps getting closer but is never truly crossed. In mathematics, an asymptote refers to a line that a graph approaches but never actually touches. For functions, asymptotic behavior describes how they behave as they move towards infinity or towards a point where they become undefined.
The problem we discussed previously illustrates this perfectly. The function \(\frac{-2}{(x-1)^{2}}\) becomes undefined at \(x=1\) because you can't divide by zero. As \(x\) leans in towards 1, from either side, the fraction's value plunges into negative infinity—which means the graph of the function drops down closer and closer towards an invisible line (the asymptote) at \(x=1\), but it’ll never reach a floor. This example helps students visualize what's happening: it’s as if the graph is being sucked into an endless pit as it nears the line \(x=1\), depicting that limitless dive towards negative infinity.
The problem we discussed previously illustrates this perfectly. The function \(\frac{-2}{(x-1)^{2}}\) becomes undefined at \(x=1\) because you can't divide by zero. As \(x\) leans in towards 1, from either side, the fraction's value plunges into negative infinity—which means the graph of the function drops down closer and closer towards an invisible line (the asymptote) at \(x=1\), but it’ll never reach a floor. This example helps students visualize what's happening: it’s as if the graph is being sucked into an endless pit as it nears the line \(x=1\), depicting that limitless dive towards negative infinity.
Calculus Problem Solving
When facing a calculus problem, a systematic approach can mean the difference between confusion and clarity. Start with a careful reading of the problem; understanding what you're being asked is half the battle. Then, identify the given information, organize your thoughts, and recall relevant formulas or theorems. If the problem involves limits, such as the one we have with the infinite limit, make sure you start by stating the definition or concept you are working with.
In our example, the process involves strategically manipulating inequalities to determine how values of \(x\) close to 1 impact the function value. At the core of calculus problem solving is the ability to break down complex behaviors into understandable parts. Here, we used a blend of algebra and conceptual reasoning to work our way through the problem, consistently keeping the definition of an infinite limit in mind.
In our example, the process involves strategically manipulating inequalities to determine how values of \(x\) close to 1 impact the function value. At the core of calculus problem solving is the ability to break down complex behaviors into understandable parts. Here, we used a blend of algebra and conceptual reasoning to work our way through the problem, consistently keeping the definition of an infinite limit in mind.
Limit Laws and Techniques
There's a toolbox of limit laws and techniques we deploy so we can handle limits systematically. Some examples include the sum, product, and quotient laws, which help simplify the limit of complex expressions. For instance, if you have the limit of a sum, you can often take the limit of each part separately.
In case of infinite limits, we often have to work with inequalities as seen in the exercise. These require a delicate touch: being mindful of direction is crucial when multiplying or dividing by negative numbers, as these actions can flip the inequality. Additionally, knowing to apply techniques such as finding common denominators, factoring, or rationalizing can transform a befuddling expression into a more manageable one. As seen in the step-by-step solution, understanding how to manipulate the function and work within the bounds of \(\delta\) and \(M\) are vital practices in solving calculus problems involving limits.
In case of infinite limits, we often have to work with inequalities as seen in the exercise. These require a delicate touch: being mindful of direction is crucial when multiplying or dividing by negative numbers, as these actions can flip the inequality. Additionally, knowing to apply techniques such as finding common denominators, factoring, or rationalizing can transform a befuddling expression into a more manageable one. As seen in the step-by-step solution, understanding how to manipulate the function and work within the bounds of \(\delta\) and \(M\) are vital practices in solving calculus problems involving limits.
Other exercises in this chapter
Problem 60
Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 0} \frac{\sin 2 x}{\sin x}$$
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Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. $$g(x)=2-\ln x^{2}$$
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Evaluate limit. $$\lim _{x \rightarrow e^{2}} \frac{\ln ^{2} x-5 \ln x+6}{\ln x-2}$$
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Transcendental functions Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of t
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