Problem 48 - Calculus Early Transcendentals

Question

We write \(\lim _{x \rightarrow a} f(x)=-\infty\) if for any negative number \(M\) there exists \(\delta>0\) such that $$f(x)

Step-by-Step Solution

Answer

Question: Prove that the limit of the function \(f(x) = \frac{-2}{(x-1)^{2}}\) as \(x\) approaches 1 is \(-\infty\). Answer: To prove that the limit as \(x\) approaches 1 is \(-\infty\), we showed that for any negative number \(M\), there exists a \(\delta\) such that whenever \(0<|x-1|<\delta\), \(f(x)=\frac{-2}{(x-1)^{2}} < M\). We found the value of \(\delta\) to be \(\delta = \sqrt{-M}\). Therefore, we can conclude that: $$\lim_{x \rightarrow 1} \frac{-2}{(x-1)^{2}} = -\infty$$

Step 1: Define the function and the limit

The function we are dealing with is \(f(x) = \frac{-2}{(x-1)^{2}}\). We want to prove that the limit as \(x\) approaches 1 is \(-\infty\): $$\lim_{x \rightarrow 1} f(x) = -\infty$$

Step 2: Use the definition of the limit

According to the definition of the limit, for any negative number \(M\), we need to find a \(\delta > 0\) such that \(f(x) < M\) whenever \(0<|x-1|<\delta\). This means, given \(M\), we need to find \(\delta\) which satisfies the following inequality: $$\frac{-2}{(x-1)^{2}} < M$$

Step 3: Manipulate the inequality

To find the \(\delta\) that satisfies the inequality, we need to manipulate the inequality as follows: $$\frac{-2}{(x-1)^{2}} < M \Rightarrow \frac{1}{(x-1)^{2}} > \frac{-1}{M}$$ Now, let's isolate \((x-1)^{2}\): $$(x-1)^{2} > \frac{1}{-\frac{1}{M}}$$ Since \(M\) is negative, \(-\frac{1}{M}\) is positive. Therefore, $$(x-1)^{2} > \frac{1}{-\frac{1}{M}} = -M$$

Step 4: Find the value of \(\delta\)

To find the value of \(\delta\), let's consider the expression \((x-1)^{2} > -M\). We know that \(0<|x-1|<\delta\). So, let's rewrite \((x-1)^{2}\) as \(|x-1|^2\): $$|x-1|^2 > -M$$ This means that, if we can find a \(\delta > 0\) such that: $$\delta^2 > -M$$ Then, the inequality will be satisfied, and we will have proven the limit. Since \(-M>0\), we can find a suitable value for \(\delta\) by taking its square root: $$\delta = \sqrt{-M}$$

Step 5: Conclusion

Using the definition of the limit, we have proven that for any negative number \(M\), there exists a \(\delta\) such that whenever \(0<|x-1|<\delta\), \(f(x)=\frac{-2}{(x-1)^{2}} < M\). We found the value of \(\delta\) to be \(\delta = \sqrt{-M}\). Therefore, we can conclude that: $$\lim_{x \rightarrow 1} \frac{-2}{(x-1)^{2}} = -\infty$$

Key Concepts

Infinite Limits

Understanding infinite limits is essential when evaluating how functions behave as their inputs approach a certain value where the function does not settle into a finite number. In simpler terms, when we say that the limit of a function is infinite as it approaches a certain point, we mean that the values of the function grow larger and larger without bound.

Thus, when dealing with infinite limits, the focus is not on finding a specific value but rather understanding this tendency to increase or decrease indefinitely. We may not observe a final 'output' value in such limits, yet studying these behaviors forms a crucial part of calculus.

To determine whether the limit is approaching positive or negative infinity, it's crucial to analyze the direction the function values take. While positive infinity entails those values growing without bounds in the positive direction, negative infinity suggests an unending decrease.

Negative Infinity

When we encounter a limit of negative infinity, it implies that as the input reaches closer to a specific value, our function plunges forever into increasingly negative values.

This is observed in the exercise through the function \( f(x) = \frac{-2}{(x-1)^2} \). Here, as \( x \) approaches 1, the denominator \((x-1)^2\) approaches zero, causing the overall fraction to produce very large negative values. This results in the limit being \(-\infty\), highlighting an ever-decreasing function behavior.

Understanding negative infinity is particularly useful in evaluating functions that may not have finite boundaries and provide deeper insights into the behavior of rational functions.

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to define the concept of a limit in calculus.

For infinite limits, as explored in our exercise, the definition tailors itself to describe how a function behaves as it approaches infinity or negative infinity. Rather than relying on two numbers, epsilon and delta, it primarily focuses on ensuring the function's behavior remains consistent with what it means to approach infinity. In this scenario, the goal is to find a distance \(\delta\) that ensures the function's values fall beyond a certain magnitude defined by \(M\), specifically making them more negative in the case of \(-\infty\).

By following this structured approach, the epsilon-delta definition helps us conclude a function's limiting behavior precisely. This gives us confidence in how functions might behave when nearing certain critical points.

Rational Functions

Rational functions, like the one in our exercise, involve quotients of polynomials. They are represented generally as \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials.

These functions can exhibit intriguing behaviors, particularly at points where their denominators equate to zero, leading to undefined values and potential for infinite limits. For instance, in our example function \( f(x) = \frac{-2}{(x-1)^2} \), the denominator approaches zero as \( x \) nears 1, causing the function's value to soar negatively as described earlier.

Understanding rational functions requires examining both the numerator and the denominator to predict the function’s behavior, especially around these undefined points. Recognizing these scenarios enables us to approximate limits effectively and harness the true potential of calculus in analyzing complex function behaviors.