Question

, find each of the right-hand and left-hand limits or state that they do not exist. $$ \lim _{x \rightarrow 3^{+}} \frac{x-3}{\sqrt{x^{2}-9}} $$

Short answer

The right-hand limit is 0.

Step-by-step solution

Understand the Problem

We need to examine the behavior of the function \( f(x) = \frac{x-3}{\sqrt{x^2-9}} \) as \( x \) approaches 3 from the right (3\(^+\)). This means we consider values of \( x \) that are slightly larger than 3 to find the right-hand limit.

Factor the Denominator

First, let's examine the denominator \( \sqrt{x^2-9} \). Notice that \( x^2-9 = (x-3)(x+3) \). As \( x \) approaches 3 from the right, the expression inside the square root approaches zero, but not equal to zero, thus \( \sqrt{x^2-9} = \sqrt{(x-3)(x+3)} \approx \sqrt{0 \times 6} = 0^+ \).

Identify the Form and Simplify

The function is of the form \( \frac{0}{0^+} \) as \( x \rightarrow 3^+ \). This is a form that suggests a type of indeterminate behavior. We simplify the expression by recognizing that as \( x \rightarrow 3^+ \), \( \frac{x-3}{\sqrt{(x-3)(x+3)}} = \frac{x-3}{\sqrt{x-3}\sqrt{x+3}} \).

Cancel Common Terms

Assuming \( x > 3 \), \( x-3 \) is positive and non-zero, which lets us cancel out \( \sqrt{x-3} \) from both the numerator and denominator:\[ \frac{x-3}{\sqrt{x-3}\sqrt{x+3}} = \frac{\sqrt{x-3}}{\sqrt{x+3}} \]

Evaluate the Simplified Limit

Simplifying gives:\( \lim_{x \to 3^+} \frac{\sqrt{x-3}}{\sqrt{x+3}} = \lim_{x \to 3^+} \frac{\sqrt{x-3}}{\sqrt{6}} \).As \( x \rightarrow 3^+ \), \( \sqrt{x-3} \rightarrow 0^+ \), so the whole expression \( \frac{0^+}{\sqrt{6}} \rightarrow 0^+ \).

Conclusion

The right-hand limit as \( x \rightarrow 3^+ \) of \( \frac{x-3}{\sqrt{x^2-9}} \) is 0.

Key concepts

Right-hand limit

A right-hand limit refers to the behavior of a function as the input (usually denoted as \(x\)) approaches a specific value from the right-hand side. This means we consider values slightly larger than the point of interest. In this context, evaluating the right-hand limit of a function as \(x\) approaches 3 from the right, noted as \(x \to 3^+\), involves taking numbers closer and closer to 3 that are slightly more than 3.

The given exercise looks at \( \lim_{x \to 3^+} \frac{x-3}{\sqrt{x^2-9}} \). To find this right-hand limit, we focus on the expression \( \frac{0}{0^+} \), indicating a form that is typically hard to evaluate plainly. Therefore, to simplify, we rewrite the expression to allow cancellation of the common terms, revealing \( \frac{\sqrt{x-3}}{\sqrt{6}} \) as \(x\) nears 3 from the right.

Calculating this, we find \( \sqrt{x-3} \) approaches \(0^+\), meaning a very small positive number divided by \(\sqrt{6}\), leaving the result as \(0^+\). Hence, this shows that the right-hand limit is \(0\).

• The right-hand limit checks the behavior from the right.
• It uses values slightly greater than the point of interest.
• Simplification often involves canceling common terms.

Left-hand limit

The left-hand limit examines the function's behavior as \(x\) approaches a specific point from the left. This approach uses values that are slightly less than the point of interest. Calculating limits from different sides helps determine the overall behavior of the function at a point.

Now, let’s explore the left-hand side of similar expressions to see what it means in a mathematical scenario. Consider the function hypothetical part \( \lim_{x \to 3^-}{\frac{x-3}{\sqrt{x^2-9}}} \). Here, we substitute values slightly less than 3 toward the analysis. This method helps identify what the function does as you approach that value from the left.

• The left-hand limit observes approach from lesser values.
• Values considered are less than the particular interest point.
• Provides insight when compared with the right-hand behavior.

Indeterminate forms

In calculus, indeterminate forms often arise when evaluating limits that initially appear undefined. The form \( \frac{0}{0} \) is a typical example indicating such occurrences. It shows that both the numerator and denominator tend to zero as \(x\) approaches a given point, making straightforward evaluation impossible. To resolve these, a deeper analysis and sometimes algebraic manipulation are required.

During your exploration of \( \lim_{x \to 3^+} \frac{x-3}{\sqrt{x^2-9}} \), we encounter the indeterminate form \( \frac{0}{0^+} \). This suggests that both parts approach a small value near zero, which otherwise would be undefined. By factoring and simplifying, the expression instead arrives at a determinate form (\( \frac{\sqrt{x-3}}{\sqrt{6}} \)), allowing for clearer evaluation.

• Indicates a need for simplification or transformation.
• Form such as \( \frac{0}{0} \), \( \frac{\infty}{fty} \), signify indeterminate limits.
• Algebraic manipulation often helps simplify these terms.