Question

Find each of the following limits or state that it does not exist. (a) \(\lim _{x \rightarrow 1} \frac{|x-1|}{x-1}\) (b) \(\lim _{x \rightarrow 1^{-}} \frac{|x-1|}{x-1}\) (c) \(\lim _{x \rightarrow 1^{-}} \frac{x^{2}-|x-1|-1}{|x-1|}\) (d) \(\lim _{x \rightarrow 1^{-}}\left[\frac{1}{x-1}-\frac{1}{|x-1|}\right]\)

Short answer

(a) DNE, (b) -1, (c) -3, (d) -∞.

Step-by-step solution

Solving part (a)

To find the limit \( \lim _{x \rightarrow 1} \frac{|x-1|}{x-1} \), note that the absolute value function \(|x-1|\) will change behavior at \(x = 1\). For \(x > 1\), \(|x-1| = x-1\), and for \(x < 1\), \(|x-1| = -(x-1)\).- As \(x \rightarrow 1^+\), the expression becomes \(\frac{x-1}{x-1} = 1 \).- As \(x \rightarrow 1^-\), the expression becomes \(\frac{-(x-1)}{x-1} = -1 \).Since the left and right limits are different, the limit does not exist.

Solving part (b)

Consider \( \lim _{x \rightarrow 1^{-}} \frac{|x-1|}{x-1} \). As \(x\) approaches 1 from the left (i.e., \(x < 1\)), \(|x-1| = -(x-1)\). Therefore, the expression simplifies to:\[ \lim _{x \rightarrow 1^{-}} \frac{-(x-1)}{x-1} = \lim _{x \rightarrow 1^{-}} -1 = -1. \]

Solving part (c)

For \( \lim _{x \rightarrow 1^{-}} \frac{x^{2}-|x-1|-1}{|x-1|} \), when \(x < 1\), \(|x-1| = -(x-1)\). Thus, the expression becomes:\[ \lim _{x \rightarrow 1^{-}} \frac{x^2 + (x-1) - 1}{-(x-1)}. \]Simplify the numerator:\[ x^2 + x - 1 - 1 = x^2 + x - 2 \]So the limit becomes:\[ \lim _{x \rightarrow 1^{-}} \frac{x^2 + x - 2}{-(x-1)}. \]Factor \(x^2 + x - 2\):\[ x^2 + x - 2 = (x-1)(x+2) \]Therefore, the expression becomes:\[ \lim _{x \rightarrow 1^{-}} \frac{(x-1)(x+2)}{-(x-1)} = \lim _{x \rightarrow 1^{-}} -(x+2) = -(1+2) = -3. \]

Solving part (d)

Consider \( \lim _{x \rightarrow 1^{-}}\left[\frac{1}{x-1}-\frac{1}{|x-1|}\right] \). When \(x < 1\), \(|x-1| = -(x-1)\), thus:\[ \lim _{x \rightarrow 1^{-}} \left[ \frac{1}{x-1} - \frac{1}{-(x-1)} \right] = \lim _{x \rightarrow 1^{-}} \left[ \frac{1}{x-1} + \frac{1}{x-1} \right]. \]Simplify:\[ 2 \times \frac{1}{x-1} = \frac{2}{x-1}. \]As \(x \rightarrow 1^{-}\), \(x-1\) approaches zero from the negative side, so \(\frac{2}{x-1}\) approaches \(-\infty\), therefore\[ \lim _{x \rightarrow 1^{-}} \left[ \frac{1}{x-1} - \frac{1}{|x-1|} \right] = -\infty. \]

Key concepts

Absolute Value

The concept of absolute value is essential when dealing with calculus limits, especially for expressions that change behavior depending on whether their arguments are positive or negative. The absolute value of a number is its distance from zero on the number line, regardless of direction. Thus, for any real number \(x\), the absolute value \(|x|\) is defined as:

• \( x \) if \( x \geq 0 \)
• \(-x \) if \( x < 0 \)

This piecewise definition is vital when calculating limits, as seen in the limit \( \lim_{x \rightarrow 1} \frac{|x-1|}{x-1} \). Here, the expression inside the absolute value, \( x-1 \), determines how \(|x-1|\) behaves as \(x\) approaches 1.
Before \(x\) equals 1, the behavior changes depending on whether \(x\) is greater than or less than 1. For \(x > 1\), \(|x-1| = x-1\), and for \(x < 1\), \(|x-1| = -(x-1)\). This switching behavior is fundamental when evaluating limits that involve absolute values.

One-sided Limits

One-sided limits allow us to evaluate the value that a function approaches as the variable approaches a particular point from only one direction – either from the left or the right. When dealing with absolute values in limit problems, understanding one-sided limits becomes crucial.
Consider the one-sided limit \( \lim_{x \rightarrow 1^{-}} \frac{|x-1|}{x-1} \). Here, \(x\) approaches 1 from the negative side or from values less than 1. This means in the expression \(|x-1|\), since \(x-1\) will be negative, the absolute value will equal the negation. Thus, the fraction becomes \( \frac{-(x-1)}{x-1} = -1 \).

• A one-sided limit from the left is denoted as \( x \rightarrow c^{-} \)
• A one-sided limit from the right is given as \( x \rightarrow c^{+} \)

One-sided limits are particularly useful for understanding the behavior of functions at points of discontinuity or where the absolute value causes a piecewise definition.

Limit Does Not Exist

The concept of a limit not existing arises when the left-hand limit and the right-hand limit at a particular point do not agree. In such cases, even though one-sided limits might exist individually, the overall limit is said to not exist.
For instance, in the problem \(\lim_{x \rightarrow 1} \frac{|x-1|}{x-1}\), we previously calculated the limits as \(x\) approaches 1 from both sides:

• From the right ( \(x \rightarrow 1^+\)): the limit is 1
• From the left ( \(x \rightarrow 1^-\)): the limit is -1

Since these two values, \(1\) and \(-1\), do not coincide, we conclude that the limit as \(x\) approaches 1 does not exist.
In general, if \( \lim_{x \rightarrow c^{+}} f(x) eq \lim_{x \rightarrow c^{-}} f(x) \), then \( \lim_{x \rightarrow c} f(x) \) is undefined or said to not exist.

Approaching Negative and Positive Sides

When evaluating limits, the direction from which \(x\) approaches the target value can affect the limit's calculation, especially when dealing with expressions that include absolute values. Understanding the distinction between approaching from the positive side and from the negative side is essential.
Approaching from the positive side of a point (\( c \)) means that you look at values of \(x\) that are greater than \( c \) (i.e., \( x \rightarrow c^+ \)). Conversely, approaching from the negative side means considering values of \(x\) that are less than \( c \) (i.e., \( x \rightarrow c^- \)).

• Approaching \(1^+\): \(x\) is slightly larger than 1
• Approaching \(1^-\): \(x\) is slightly smaller than 1

In the examined exercise, these distinctions play crucial roles. For example, in part (d), calculating \( \lim_{x \rightarrow 1^{-}} \left[ \frac{1}{x-1} - \frac{1}{|x-1|} \right] \) requires understanding how each component behaves as \(x\) approaches 1 from the negative side. Expression \( \frac{1}{x-1} \) becomes more negative, suggesting the limit approaches \(-\infty\) because as \(x\) gets infinitesimally close to 1 from the left, it makes the denominator approach a negative zero.