Question

\(\operatorname{Lt}_{x \rightarrow \frac{2}{5}} \frac{1}{|5 x-2|}=\) (1) 0 (2) \(\infty\) (3) 1 (4) Does not exist

Short answer

Answer: (4) Does not exist

Step-by-step solution

Check Continuity

First, let's express the function in a piecewise form, as the absolute value can be described as two different signs. We can write the absolute value as \(|5x - 2| = (5x - 2)\) when \(5x - 2 \geq 0\) and \(|5x - 2| = -(5x - 2)\) when \(5x - 2 < 0\). So our function is: $$ \frac{1}{|5x-2|} = \begin{cases} \frac{1}{5x-2} & \text{if } x \geq \frac{2}{5} \\ -\frac{1}{5x-2} & \text{if } x < \frac{2}{5} \end{cases} $$ Notice that our function is continuous on either side of the point \(x = \frac{2}{5}\).

Calculating the Limit

Now let's find the limit on either side of \(x = \frac{2}{5}\). Calculate the limit as \(x\) approaches \(\frac{2}{5}\) from the left side. $$ \lim_{x \to \frac{2}{5}^{-}} \frac{1}{|5x-2|} = \lim_{x \to \frac{2}{5}^{-}} -\frac{1}{5x-2} $$ As \(x \to \frac{2}{5}^{-}\), \(5x-2 \to 0^{-}\). Therefore, the limit as \(x \to \frac{2}{5}^{-}\) is: $$ \lim_{x \to \frac{2}{5}^{-}} -\frac{1}{5x-2} = -\infty $$ Now, calculate the limit as \(x\) approaches \(\frac{2}{5}\) from the right side. $$ \lim_{x \to \frac{2}{5}^{+}} \frac{1}{|5x-2|} = \lim_{x \to \frac{2}{5}^{+}} \frac{1}{5x-2} $$ As \(x \to \frac{2}{5}^{+}\), \(5x-2 \to 0^{+}\). Therefore, the limit as \(x \to \frac{2}{5}^{+}\) is: $$ \lim_{x \to \frac{2}{5}^{+}} \frac{1}{5x-2} = \infty $$

Conclusion

Since the limit is different when approaching the point from the left side and the right side, we can conclude that the limit does not exist. Therefore, the correct answer is: (4) Does not exist

Key concepts

Continuity in Functions

In calculus, continuity is an important concept that ensures a function behaves predictably. A function is said to be continuous at a point if three main conditions are satisfied: the function is defined at the point, the limit of the function exists as the point is approached from both sides, and the value of the function at that point is equal to the limit. In simpler terms, a continuous function does not "break" or exhibit any abrupt changes or jumps.
When analyzing the function \[ \frac{1}{|5x-2|} \]for continuity, it is essential to consider the function without the influence of absolute value first. The presence of the absolute value \(|5x-2|\)requires us to handle it differently for various ranges of \(x\). Despite being continuous on intervals not including the critical point \(x = \frac{2}{5}\), the point itself is not continuous due to differing behavior as you approach from either side. The breakdown in continuity is linked to the change in the function's structure dictated by the absolute value.

Piecewise Functions

Piecewise functions are essential in calculus as they allow the definition of a function in sections, each with specific rules or formulas. They are particularly useful when a single mathematical expression cannot wholly define a function across its entire domain. For the function in question:\[\frac{1}{|5x-2|} = \begin{cases}\frac{1}{5x-2} & \text{if } x \geq \frac{2}{5} \-\frac{1}{5x-2} & \text{if } x < \frac{2}{5}\end{cases}\]The absolute value \(|5x-2|\)leads us to describe the function piecewise. This indicates that there are two distinct expressions based on the position of \(x\) relative to \(\frac{2}{5}\). Such an arrangement allows handling the otherwise complex behavior of the function systematically and clearly. Each segment of the function can be examined independently over its defined interval, contributing to our understanding of how the function behaves differently on either side of a specific point.

One-Sided Limits

One-sided limits are key concepts when working with pieces of functions with discontinuities. Unlike regular limits that approach a point from both sides, one-sided limits examine the behavior of a function as it approaches a point from only one direction—either left or right.
In our exercise, calculating the limit of\[ \frac{1}{|5x-2|} \]as \(x\) approaches \(\frac{2}{5}\) requires considering one-sided limits:

• From the left: \( \lim_{x \to \frac{2}{5}^{-}} -\frac{1}{5x-2} = -\infty \)
• From the right: \( \lim_{x \to \frac{2}{5}^{+}} \frac{1}{5x-2} = \infty \)

The limits from both directions differ, resulting in the overall limit not existing at that point. Recognizing and calculating one-sided limits provides valuable insight for understanding discontinuities and behavior at specific points of piecewise functions. These analyses make it easier to predict and explain abrupt changes like jumps or unbounded behavior near discontinuities. Understanding these concepts enables a thorough grasp of function behavior around points where typical evaluations might break down.