Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 1^{+}} \frac{x-1}{\sqrt{x^{2}-1}}$$

Short answer

Answer: The limit of the function as x approaches 1 from the positive side is \(\frac{1}{2}\).

Step-by-step solution

Identify the limit

The given function is \(\frac{x-1}{\sqrt{x^2 - 1}}\). We are supposed to find the limit of this function as \(x\) approaches \(1\) from the positive side. We write the given limit as: $$\lim _{x \rightarrow 1^{+}} \frac{x-1}{\sqrt{x^{2}-1}}$$

Check the denominator for a problem

Let's check what happens with the denominator as \(x\) approaches 1. If the denominator approaches zero or undefined, then the limit might not exist. Consider the denominator for the given function: $$\sqrt{x^2 - 1}$$ As x approaches \(1^{+}\), the denominator would become \(\sqrt{1^2 - 1}\), which is equal to \(0\). This means that the limit of the expression may not exist, as the denominator approaches zero.

Perform the simplification

Simplify the given function by multiplying and dividing by the conjugate of the denominator to see if any simplifications can be done. Multiply both the numerator and denominator by \(\sqrt{x^2 - 1} + 1\): $$\lim _{x \rightarrow 1^{+}}\frac{(x-1)(\sqrt{x^2-1}+1)}{(\sqrt{x^{2}-1})(\sqrt{x^{2}-1}+1)}$$ Expanding the denominator gives \((x^2 - 1)(\sqrt{x^2 -1} + 1) - (\sqrt{x^2 -1} + 1)\). After simplifying (as the middle term will be zero), we get: $$\lim _{x \rightarrow 1^{+}}\frac{(x-1)(\sqrt{x^2-1}+1)}{x^2 - 1}$$

Factor the expression

Factor the simplified expression: $$\lim _{x \rightarrow 1^{+}}\frac{(x-1)(\sqrt{x^2-1}+1)}{(x-1)(x+1)}$$

Cancel the common factors

Observe that both the numerator and the denominator have a common factor \((x-1)\). Cancel this factor out to get: $$\lim _{x \rightarrow 1^{+}}\frac{\sqrt{x^2-1}+1}{x+1}$$

Evaluate the limit

Now we can evaluate the limit as \(x\) approaches \(1^+\): $$\lim _{x \rightarrow 1^{+}}\frac{\sqrt{x^2-1}+1}{x+1} = \frac{\sqrt{1^2-1}+1}{1+1} = \frac{0+1}{2}$$ Therefore, the limit as x approaches 1 from the positive side is equal to \(\frac{1}{2}\).

Key concepts

Limit Evaluation

When solving calculus problems, evaluating limits is crucial. It helps determine the behavior of functions as they approach particular points. In simple terms, limits help us **understand what value a function is approaching** when the independent variable gets **very close but never actually reaches a certain point**. This is especially important in functions that may not be well-defined at exactly that point.

To evaluate limits, follow these general steps:

• Identify the limit expression: Determine the function whose limit you are trying to find and the point where the variable is approaching.
• Check direct substitution: See if you can directly substitute the value of the variable into the function. If this works without any issues, you have your limit. If you encounter an indeterminate form (like \(\frac{0}{0}\)), proceed to the next step.
• Simplify and factor: If direct substitution results in indeterminate forms, try to simplify the expression. This could involve factoring, rationalizing the denominator, or rewriting the expression to eliminate the issue.
• Reevaluate the limit: After simplifying, try substituting again or use other limit properties or techniques if necessary.

Understanding and identifying limits allow us to smoothly transition through calculus concepts and solve both simple and complex problems.

Function Simplification

In calculus, simplifying a function often makes limit evaluation more manageable. Simplification is the process of transforming a function into a form that is easier to work with, especially when direct substitution doesn't immediately resolve the limit.

When simplifying a function for limits:

• Identify obstacles: Look for elements that complicate direct evaluation—such as terms resulting in division by zero or undefined operations.
• Use conjugates: Conjugates are particularly helpful with irrational expressions. By multiplying by the conjugate, terms like square roots can often be eliminated or simplified.
• Factor expressions: Factor common terms and use them to simplify the expression. This often results in cancelling terms that cause indeterminate forms.
• Rewriting expressions: Sometimes rewriting a term can clarify its behavior. For example, turning a quotient into a product, or splitting terms.
• Simplify step-by-step: Always work through simplifications in a systematic manner to prevent errors.

Simplification not only facilitates the evaluation of limits but also provides insights into the behavior of various kinds of functions.

Directional Limits

Directional limits are a special focus when examining functions at points where behavior changes. Unlike two-sided limits, directional limits consider a function's approach from only one side — either from the left or the right.

To understand directional limits, consider:

• One-sided approach: When calculating a limit like \(\lim_{x \to a^+} f(x)\), you're looking at how the function behaves as \(x\) approaches \(a\) from the right (positive side). Conversely, \(\lim_{x \to a^-} f(x)\) considers the function from the left (negative side).
• Behavior at discontinuities: Directional limits allow us to examine functions at points where they might not be continuous, such as step functions or functions with sharp turns.
• Existence and equality: For a two-sided limit to exist at a certain point, both directional limits (from the left and right) need to exist and be equal. If they're not equal, the two-sided limit does not exist at that point.

Directional limits give insights into the localized behavior of functions and are essential when evaluating functions with discontinuities or abrupt changes in direction. They help us fully grasp how a function behaves near critical points, providing a clearer view of its continuity.