Problem 49 - Calculus Early Transcendentals

Question

Evaluate the following limits, where a and \(b\) are fixed real numbers. \(\lim _{x \rightarrow a} \frac{x-a}{\sqrt{x}-\sqrt{a}}, a>0\)

Step-by-Step Solution

Answer

Question: Evaluate the limit of the following expression as x approaches a, where a is a positive real number: $$\lim _{x \rightarrow a} \frac{x-a}{\sqrt{x}-\sqrt{a}}$$ Answer: The limit of the given expression as x approaches a is \(2\sqrt{a}\).

Step 1: Rewrite the limit

Rewrite the limit using the given expression: $$\lim _{x \rightarrow a} \frac{x-a}{\sqrt{x}-\sqrt{a}}$$

Step 2: Rationalize the expression

In order to handle the indeterminate form, we will multiply the numerator and denominator by the conjugate of the denominator: $$\lim _{x \rightarrow a} \frac{x-a}{\sqrt{x}-\sqrt{a}} \cdot \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}$$

Step 3: Simplify the expression

Now, multiply the expression and remember that \((\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a}) = x-a\) by using the difference of squares: $$\lim _{x \rightarrow a} \frac{(x-a)(\sqrt{x}+\sqrt{a})}{x-a}$$

Step 4: Cancel the common factors

The \((x-a)\) factors in the numerator and denominator cancel each other: $$\lim _{x \rightarrow a} \sqrt{x}+\sqrt{a}$$

Step 5: Evaluate the limit

Now that the expression has been simplified, we can evaluate the limit by directly substituting \(x=a\): $$\sqrt{a}+\sqrt{a} = 2\sqrt{a}$$ Thus, the limit is equal to \(2\sqrt{a}\).

Key Concepts

Evaluating Limits

In calculus, evaluating limits is a fundamental process that investigates what happens to a function as the input approaches a specific value, but never actually reaches it. It's a way to understand behavior at points where direct substitution may not give clear answers. When you evaluate the limit of a function such as \( \lim _{x \rightarrow a} \frac{x-a}{\sqrt{x}-\sqrt{a}} \) where \(a>0\), you're looking for what value the function settles on as \(x\) approaches \(a\).

Rationalizing the numerator or denominator is often an essential step to simplify the process of evaluating limits, especially when indeterminate forms like \(0/0\) arise. By multiplying by conjugates and using algebraic rules like the difference of squares, we simplify expressions and pave the way for direct substitution or further analysis.

Rationalizing Expressions

To deal with complex expressions, especially those involving square roots, rationalizing expressions is a useful technique. This involves multiplying the expression by a quantity known as the 'conjugate' to eliminate irrational numbers or variables from the denominator of a fraction. The conjugate of \(\sqrt{x}-\sqrt{a}\) is \(\sqrt{x}+\sqrt{a}\), and when we multiply them, the product leverages the difference of squares formula. By doing this, not only do we rationalize the expression, but we also often resolve indeterminate forms encountered during limit evaluation, paving the way for a simpler expression that can be easily handled.

Difference of Squares

The difference of squares is an algebraic pattern that comes into play when we have an expression of the form \(a^2-b^2\). The key is to recognize that this expression can be factored into \(a-b\) times \(a+b\). In our limit problem, after multiplying the numerator and denominator by the conjugate, we apply the difference of squares to get from \(\sqrt{x}^2 - \sqrt{a}^2\) to \(x-a\). This step is crucial in simplifying the expression and getting closer to evaluating the limit. Many calculus problems rely on being able to recognize and apply this pattern to move forward in the solution process.

Indeterminate Forms

In calculus, an indeterminate form is an expression that doesn’t offer immediate insight into the behavior of a function as an input approaches some limit. The most common indeterminate form is \(0/0\), but others include \(\infty/\infty\), \(0 \cdot \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\). These forms cannot be evaluated through direct substitution; instead, they call for algebraic manipulation, like rationalizing expressions, or analytical tools such as L'Hôpital's rule, to find a determinate value. In our limit problem, we used rationalizing to move past the \(0/0\) indeterminate form and successfully evaluate the limit.