Problem 19 - Calculus: Early Transcendentals

Question

Which of the following statements are correct? Choose all that apply. a. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}\) does not exist \(\quad\) b. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=\infty\) c. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=-\infty\)

Step-by-Step Solution

Answer

a. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}\) does not exist. b. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=\infty\). c. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=-\infty\). Answer: a, b

Step 1: Identify the function and the limit point

We are given the function \(f(x) = \frac{1}{(x-1)^{2}}\) and we want to analyze the limit as \(x\) approaches \(1\).

Step 2: Analyze the function near x=1

When \(x\) is close to \(1\) but not equal to \(1\), the denominator \((x-1)^2\) is very close to \(0\). Since the denominator is squared, it will always be positive. As \(x\) approaches \(1\), the denominator becomes smaller and smaller, making the whole fraction \(\frac{1}{(x-1)^{2}}\) larger and larger.

Step 3: Determine the correct statements#a. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}\) does not exist

Since the function grows without bound as \(x\) approaches \(1\), the limit does not exist in the traditional sense (i.e., it does not approach a finite value). So, this statement is correct.

Step 4: Determine the correct statements#b. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=\infty\)

As we analyzed in step 2, the function grows without bound as \(x\) approaches \(1\). Hence, the limit is infinite. So, this statement is also correct.

Step 5: Determine the correct statements#c. \(\lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=-\infty\)

Since the denominator \((x-1)^2\) is always positive, the function \(f(x)\) is also always positive. As \(x\) approaches \(1\), the function grows without bound in the positive direction. So, the limit is not negative infinity. Hence, this statement is incorrect. In conclusion, statements a and b are correct while statement c is incorrect.

Key Concepts

Limits and Infinity

Understanding the concept of limits as they approach infinity is crucial for grasping the behavior of functions near points where they do not remain bounded. In the original exercise, we observed the limit of the function \( f(x) = \frac{1}{(x-1)^{2}} \) as \( x \) approaches 1. What we notice is that as \( x \) gets closer to 1, the value of \( f(x) \) becomes larger and larger, essentially growing without bound.

This unbounded growth is what we represent by the notation \( \lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}}=\infty \). It expresses that the limit does not exist in the traditional sense since it does not approach a fixed numerical value but rather increases indefinitely. When dealing with limits and infinity, it's important to recognize the direction of growth (positive or negative infinity) and how the function behaves just before reaching the point of interest. In this case, the function exhibits a positive growth towards infinity as \( x \) approaches 1.

Analyzing Functions

Analyzing functions is a fundamental skill in calculus that involves examining the behavior of a function at various points, especially concerning its limits, continuity, and differentiability. The given example requires analysis around the point \( x = 1 \) for the function \( f(x) = \frac{1}{(x-1)^{2}} \).

To thoroughly analyze this function, we consider what happens as we approach the value of 1 from both the left and the right. Despite the approach direction, the denominator \( (x-1)^2 \) always remains positive, eliminating the possibility of a negative function value. As a result, we determine that the function's trend is to infinitely increase as \( x \) nears 1, indicating a discontinuity at this point since the function does not yield a real number value at \( x = 1 \) but instead trends towards positive infinity.

Limits at Points of Discontinuity

When approaching points of discontinuity, the behavior of a function near these points is often non-intuitive and requires careful analysis. In our exercise, the limit in question demonstrates a common type of discontinuity, where \( f(x) \) becomes infinitely large as it approaches a specific \( x \) value, in this case, 1.

It's essential to distinguish this from other types of discontinuity, such as a jump or removable discontinuity. This particular discontinuity, where a function approaches infinity, is called an infinite discontinuity. Here, \( \lim _{x \rightarrow 1} \frac{1}{(x-1)^{2}} \) does not approach a finite value, and is instead said to diverge to infinity. The exercise improvement advice comes in handy when identifying this behavior: by looking at how the function behaves just before and after the point of discontinuity and noting the sign of the function's output, we conclude that although the limit 'does not exist' in the traditional sense, it does approach infinity in a definite, positive direction.