Problem 57 - Thomas Calculus

Question

Let $$ f(x)=\left\\{\begin{array}{ll}{x,} & {x<1} \\ {x+1,} & {x>1}\end{array}\right.$$ $$ \begin{array}{l}{\text { a. Let } \epsilon=1 / 2 . \text { Show that no possible } \delta>0 \text { satisfies the fol- }} \\ {\text { lowing condition: }}\end{array} $$ $$x, \quad 0<|x-1|<\delta \Rightarrow|f(x)-2|<1 / 2$$ That is, for each \(\delta>0\) show that there is a value of \(x\) suchthat $$0<|x-1|<\delta \quad \text { and } \quad|f(x)-2| \geq 1 / 2$$ This will show that \(\lim _{x \rightarrow 1} f(x) \neq 2\) $$\begin{array}{l}{\text { b. Show that } \lim _{x \rightarrow 1} f(x) \neq 1} \\\ {\text { c. Show that } \lim _{x \rightarrow 1} f(x) \neq 1.5}\end{array}$$

Step-by-Step Solution

Answer

The limit \( \lim_{x \to 1} f(x) \) is undefined because there is a jump discontinuity at \( x = 1 \).

Step 1: Understanding the Problem

We need to examine the limit of the function \( f(x) \) as \( x \) approaches 1. Our task is to show that the function does not approach specific values, given the discontinuity or jump at \( x = 1 \). We will apply the epsilon-delta criterion for limits to prove these points.

Step 2: Part a: Initial Setup

We are tasked to show that there's no \( \delta > 0 \) such that the condition \( 0 < |x-1| < \delta \) implies \( |f(x) - 2| < \frac{1}{2} \). The function is defined as \( f(x) = x \) if \( x < 1 \) and \( f(x) = x + 1 \) if \( x > 1 \). Check specifically near \( x = 1 \).

Step 3: Part a: Analyze \( x < 1 \)

For \( x < 1 \), \( f(x) = x \). Therefore, \( |f(x) - 2| = |x - 2| \). If we check a small \( x < 1 \), say \( x = 0.5 \), \( |0.5 - 2| = 1.5 \geq 0.5 \). Hence, no \( \delta \) satisfies the condition for every \( x < 1 \).

Step 4: Part a: Analyze \( x > 1 \)

For \( x > 1 \), \( f(x) = x + 1 \). Therefore, \( |f(x) - 2| = |x + 1 - 2| = |x - 1| \). Choose \( x = 1.5 \). Then \( |1.5 - 1| = 0.5 \geq 0.5 \), which violates the condition. Thus, no \( \delta \) satisfies this either.

Step 5: Part a: Conclusion

Since there does not exist a \( \delta > 0 \) such that the condition is satisfied for all \( x \) in the neighborhood of 1, except \( x = 1 \), we demonstrate that \( \lim_{x \to 1} f(x) eq 2 \).

Step 6: Part b: Verify \( \lim_{x \to 1} f(x) \neq 1 \)

Using the same approach, check if \( |f(x) - 1| < \epsilon \) for some \( \epsilon > 0 \). For \( x \to 1^+ \), \( f(x) = x + 1 \rightarrow 2 \). For \( x \to 1^- \), \( f(x) = x \rightarrow 1 \). Since both one-sided limits do not agree, \( \lim_{x \to 1} f(x) eq 1 \).

Step 7: Part c: Verify \( \lim_{x \to 1} f(x) \neq 1.5 \)

Now test \( |f(x) - 1.5| < \epsilon \). For \( x < 1 \), \( f(x) = x \rightarrow 1 \) which gives \( |1 - 1.5| = 0.5 \). For \( x > 1 \), \( f(x) = x + 1 \rightarrow 2 \), giving \( |2 - 1.5| = 0.5 \). Since either direction violates the condition for a small \( \delta \), \( \lim_{x \to 1} f(x) eq 1.5 \).

Step 8: Final Conclusion

The limit of \( f(x) \) as \( x \to 1 \) is undefined because the function's left-hand and right-hand limits do not agree, thus it doesn't approach any specific value among those specified (1, 1.5, or 2).

Key Concepts

Discontinuity

In mathematics, discontinuity occurs when a function does not connect smoothly in a particular point within its domain. In other words, if you imagine drawing a graph of the function without lifting your pen, a discontinuity would be a spot where you'd have to lift your pen. In the given exercise, the function \( f(x) = \begin{cases} x & \text{if } x < 1 \ x + 1 & \text{if } x > 1 \end{cases} \)exhibits a discontinuity at \( x = 1 \). - When \( x < 1 \), the function simply equals \( x \), and for \( x > 1 \), it jumps to \( x + 1 \). - This jump means that there is a sudden change in the value of the function at this point. - For example, if you approach \( x = 1 \) from the left (such as at \( x = 0.9 \)), the function outputs \( x \). From the right (such as at \( x = 1.1 \)), it outputs \( x+1 \). This disparity in values when approaching from different sides confirms the discontinuity at \( x = 1 \). Discontinuities can greatly affect limits since limits rely on the function's behavior to be seamless or continuous.

Epsilon-Delta Criterion

The epsilon-delta criterion is a formal definition of the limit in calculus. It establishes how close a function should be to a certain limit value as the input gets arbitrarily close to a specified point. - For a function \( f(x) \), we write \( \lim_{x \to a} f(x) = L \) if for every real number \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x-a| < \delta \), it follows that \( |f(x) - L| < \epsilon \). Using this criterion, we analyze limits by comparing desired closeness (given by \( \epsilon \)) with input proximity (controlled by \( \delta \)). In the problem, we were tasked with showing that no matter how small a \( \delta \) we pick, we cannot ensure \( |f(x) - 2| < \frac{1}{2} \) near \( x = 1 \). This demonstrates the failure of \( f(x) \) to approach the value 2 as \( x \to 1 \), illustrating that the limit does not exist at this discontinuity, failing the epsilon-delta condition.

One-sided limits

One-sided limits help in understanding a function's behavior by considering the approach of inputs from one side only. This is especially useful in identifying discontinuities or jumps in functions. When discussing a limit, it can be taken as a one-sided from the left or the right. - A left-hand limit, noted as \( \lim_{x \to a^-} f(x) \), refers to approaching a value \( a \) from values less than \( a \). - Conversely, a right-hand limit, noted as \( \lim_{x \to a^+} f(x) \), considers approaching \( a \) from values greater than \( a \).In our scenario with the function:- The left-hand limit as \( x \to 1^- \) (from the left) gives a value of \( f(x) = x \), tending towards 1. - The right-hand limit \( x \to 1^+ \) (from the right) evaluates to \( f(x) = x + 1 \), trending towards 2.The discrepancy between these one-sided limits means that the overall limit \( \lim_{x \to 1} f(x) \) does not exist. One-sided limits are often key in pointing out where a function fails to have a regular limit, highlighting discontinuities.