Problem 46 - Thomas Calculus

Question

Find the limits in Exercises \(37-48.\) $$\text { a. }\lim _{x \rightarrow 0^{+}} \frac{2}{x^{1 / 5}} \quad \text { b. } \lim _{x \rightarrow 0^{-}} \frac{2}{x^{1 / 5}}$$

Step-by-Step Solution

Answer

a. \(+\infty\); b. \(-\infty\).

Step 1: Understand the Problem

We need to find two limits as \(x\) approaches 0 from the positive direction \((0^+)\) and from the negative direction \((0^-)\) for the function \( f(x) = \frac{2}{x^{1/5}} \). This means we will explore the behavior of \( f(x) \) as \( x \to 0 \) from both sides.

Step 2: Analyze the Function

The function is \( f(x) = \frac{2}{x^{1/5}} \). Since the denominator has a fractional power, it behaves differently for positive and negative values of \(x\). Remember that the 5th root of \(x\) when positive remains positive, and when negative, it also remains negative.

Step 3: Calculate the Limit from Right

Consider \(\lim_{x \to 0^+} \frac{2}{x^{1/5}}\). As \(x\) approaches 0 from the right (positive values), \(x^{1/5}\) becomes a very small positive number. The fraction \( \frac{2}{x^{1/5}} \) tends towards infinity since you are dividing by a number that decreases close to zero. Therefore, the limit is \[ \lim_{x \to 0^+} \frac{2}{x^{1/5}} = +\infty \]

Step 4: Calculate the Limit from Left

Now consider \(\lim_{x \to 0^-} \frac{2}{x^{1/5}}\). As \(x\) approaches 0 from the left (negative values), \(x^{1/5}\) becomes a very small negative number because the 5th root of a negative number remains negative. Thus, \( \frac{2}{x^{1/5}} \) tends towards negative infinity as you are dividing by a number that is negative and close to zero. Therefore, the limit is \[ \lim_{x \to 0^-} \frac{2}{x^{1/5}} = -\infty \]

Key Concepts

One-Sided Limits

One-sided limits involve evaluating the behavior of a function as the variable approaches a specific value from only one side — either the left or the right. In our exercise, we explored the limits as \( x \) approaches zero from the positive direction \((0^+)\) and from the negative direction \((0^-)\).

When we talk about \( \lim_{x \to 0^+} \), we're interested in what happens to the function as \( x \) gets closer and closer to zero through values that are greater than zero. Conversely, \( \lim_{x \to 0^-} \) considers \( x \) approaching zero through negative values.

• For \( \lim_{x \to 0^+} \), the focus is on values just slightly larger than zero, where we see that the function grows infinitely larger.
• For \( \lim_{x \to 0^-} \), the focus switches to values slightly smaller than zero, where the function moves towards -∞.

These one-sided limits highlight how a function can behave differently based on the direction from which \( x \) approaches the limit point.

Fractional Exponents

Fractional exponents, like \( x^{1/5} \) in this exercise, represent roots of numbers. A fractional exponent \( \frac{1}{n} \) is equivalent to taking the nth root of a number. For instance, \( x^{1/5} \) is the same as saying the fifth root of \( x \).

Here's the thing about fractional exponents:

• They can be applied to both positive and negative numbers.
• A negative base with an odd root (like the 5th root) will result in a negative number.
• Consider \( x^{1/5} \): For positive \( x \), \( x^{1/5} \) is positive. For negative \( x \), \( x^{1/5} \) remains negative.

In our limit problem, the nature of the fractional exponent impacts the behavior of the function depending on whether \( x \) is positive \( (0^+) \) or negative \( (0^-) \), leading to different limit results.

Infinity in Limits

Infinity often appears in limit problems where the function grows without bound as the variable approaches a particular point. In the given exercise, both limits resulted in either \(+\infty\) or \(-\infty\), depending on whether \( x \) approaches 0 from the right or from the left.

Understanding how a function approaches infinity is crucial:

• For \( \lim_{x \to 0^+} \), \( \frac{2}{x^{1/5}} \) tends towards \(+\infty\) because the very small positive denominator causes the fraction to increase without any bounds.
• For \( \lim_{x \to 0^-} \), the denominator \( x^{1/5} \) is a very small negative value, causing the function to head towards \(-\infty\).

Limits involving infinity help us understand the unbounded behavior of functions as they approach certain critical points. This is especially useful in realms like physics and engineering, where such behavior is often observed in real-world problems.