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Find each limit, if it exists: (a) \(\lim _{x \rightarrow a^{-}} f(x)\) (b) \(\lim _{x \rightarrow a^{+}} f(x)\) (c) \(\lim _{x \rightarrow a} f(x)\) $$ f(x)=x^{2 / 3} ; \quad a=-8 $$

Short Answer

Expert verified
Both one-sided limits and the two-sided limit are 4.

Step by step solution

01

Analyze the Function

Consider the function \(f(x) = x^{2/3}\). This function is defined for all real numbers because the exponent \(2/3\) means we take the cube root of \(x\) and square it, which is possible for negative and positive \(x\).
02

Find the Left-Hand Limit

To find \(\lim_{x \to -8^-} f(x)\), consider values of \(x\) approaching \(-8\) from the left. For these values, \(x = -8 - \epsilon\) where \(\epsilon\) is a small positive number, leading \(f(x) = ((-8 - \epsilon)^{1/3})^2\). Since cube roots of negative numbers are negative, \((-8 - \epsilon)^{1/3}\) is negative, but squaring makes \(f(x)\) positive. As \(x\) approaches \(-8\), the value approaches \((-8)^{1/3})^2 = 4\). Thus, \(\lim_{x \to -8^-} f(x) = 4\).
03

Find the Right-Hand Limit

For \(\lim_{x \to -8^+} f(x)\), consider values of \(x\) approaching from the right of \(-8\). Here, \(x = -8 + \epsilon\), which results in \(f(x) = ((-8 + \epsilon)^{1/3})^2\). The behavior is similar as before: the cube root of a value slightly larger than negative is still negative, but squaring gives \(((-8 + \epsilon)^{1/3})^2 = ((-8)^{1/3})^2 = 4\). Thus, \(\lim_{x \to -8^+} f(x) = 4\).
04

Determine the Two-Sided Limit

Since both the left-hand limit and right-hand limit as \(x\) approaches \(-8\) are equal to 4, the two-sided limit exists. Therefore, \(\lim_{x \to -8} f(x) = 4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Left-Hand Limit
When examining limits in calculus, a left-hand limit (also known as a "limit from the left") focuses on how a function behaves as the variable approaches a specific value from the left side. In more technical terms, for a function \(f(x)\) and a point \(a\), we denote the left-hand limit as \(\lim_{x \to a^-} f(x)\). Here, the minus sign indicates we are considering values of \(x\) slightly less than \(a\).

In the given exercise, we look at \(\lim_{x \to -8^-} f(x)\). These are values of \(x\) approaching a negative 8 from the left. Practically, it means we're considering numbers like -8.1, -8.01, -8.001, and so on. With the function \(f(x) = x^{2/3}\), even though \(x\) is negative, eventually, we take the cube root (which outputs a negative number) and then square it, resulting in a positive value.

Thus, for this function, as \(x\) values get closer to -8 from the left, \(f(x)\) consistently approaches 4. The left-hand limit, therefore, is 4.
Right-Hand Limit
Understanding right-hand limits, or "limits from the right," is similar to left-hand limits, but we look at the behavior of a function as it approaches from greater values. To express this, we use \(\lim_{x \to a^+} f(x)\), where the plus sign signifies approaching \(a\) from the right.

In our practice problem, analyze \(\lim_{x \to -8^+} f(x)\). Here, we examine \(x\) as it approaches negative 8 from slightly larger values. Think of examples like -7.9, -7.99, -7.999, closing in on -8. With \(f(x) = x^{2/3}\), when \(x\) is barely larger than -8, the cube root produces a negative result, but the squaring process turns it back positive.

This behavior ensures that just like the left-hand limit, the right-hand result approaches 4. Hence, as \(x\) approaches -8 from the right, the right-hand limit is found to also be 4.
Two-Sided Limit
The concept of a two-sided limit revolves around understanding what value a function approaches as the input comes from both directions—both from the left and the right. This is expressed as \(\lim_{x \to a} f(x)\). For this type of limit to exist, the left-hand limit must equal the right-hand limit at the point \(a\).

In this problem, after computing both left and right-hand limits as \(x\) approaches -8, we determined both were equal to 4. This equality confirms the existence of a two-sided limit, allowing us to conclusively state \(\lim_{x \to -8} f(x) = 4\).

In this scenario, because both one-sided limits matched, it showed the function's behavior was consistent as \(x\) neared -8 from any direction. We can then confidently say the two-sided limit confirms the function value the input approaches is truly 4.

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