Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9}$$

Short answer

Answer: The limit of the given function as x approaches 9 is 1.

Step-by-step solution

Identify the limit expression

The given limit expression is: $$\lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9}$$

Find the common factor

Observe that when \(x=9\), both the numerator and denominator become zero. This suggests that there must be a common factor. We can rewrite the numerator as follows: \(\sqrt{x} - 3 = \sqrt{x} - \sqrt{9}\) Now, applying the difference of squares factorization: \(\sqrt{x} - \sqrt{9} = (\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})\)

Rewrite the function with the identified factor

Using the factorization from the previous step, we rewrite the function as: $$\lim _{x \rightarrow 9} \frac{(\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})}{x-9}$$ However, we can also rewrite the denominator \(x-9\) as \((\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})\). Therefore, the function becomes: $$\lim _{x \rightarrow 9} \frac{(\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})}{(\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})}$$

Simplify the function before evaluating the limit

Now we can cancel out the common factors of the numerator and the denominator: $$\lim _{x \rightarrow 9} \frac{(\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})}{(\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})} = \lim _{x \rightarrow 9} \frac{\cancel{(\sqrt{x}+\sqrt{9})}(\sqrt{x}-\sqrt{9})}{\cancel{(\sqrt{x}+\sqrt{9})}(\sqrt{x}-\sqrt{9})}$$ Which simplifies to: $$\lim _{x \rightarrow 9} \frac{\sqrt{x}-\sqrt{9}}{\sqrt{x}-\sqrt{9}} $$ As we have removed the common factor that caused the denominator to become zero, we can now safely evaluate the limit.

Evaluate the simplified limit

Now that we have simplified the function, we can evaluate the limit as x approaches 9: $$\lim _{x \rightarrow 9} \frac{\sqrt{x}-\sqrt{9}}{\sqrt{x}-\sqrt{9}} = \frac{\sqrt{9}-\sqrt{9}}{\sqrt{9}-\sqrt{9}} = \frac{0}{0}$$ However, since the expression we have is a constant, dividing 0 by 0 in this context yields 1: $$\lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9} = 1$$ The limit, as x approaches 9, of the given function is 1.

Key concepts

Difference of Squares

The difference of squares is an essential algebraic identity that helps when dealing with expressions containing square roots and can be key to resolving indeterminate forms like \(\frac{0}{0}\). This identity is expressed as \(a^2 - b^2 = (a-b)(a+b)\). When it comes to functions involving limits, recognizing an expression as a difference of squares can allow us to factor and simplify it.

For example, in the original exercise, \(\sqrt{x} - \sqrt{9}\) was identified as a difference of squares and rewritten using the identity as \(\sqrt{x} - \sqrt{9} = (\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})\). By doing this, we make the expression easier to handle and manage the indeterminate form caused by the zeros in the numerator and denominator. Factorization like this can cancel out instances of \(\frac{0}{0}\), helping you find or simplifying complex expressions into something more manageable.

Simplifying Expressions

Simplifying expressions in the context of limit calculations is about making the problem more manageable by reducing it to its simplest form. This process often involves factoring, expanding, or eliminating terms, making it easier to compute the limit as a variable approaches a specific value.

In the exercise, we simplified the complex fraction by recognizing the common factor in the numerator and the denominator. By rewriting it using the difference of squares, we were able to simplify the expression effectively: \((\sqrt{x}+\sqrt{9})(\sqrt{x}-\sqrt{9})\) in both the numerator and the denominator of the limit expression canceled each other out. This simplification is crucial because it resolves the indeterminate form \(()\) that occurs when both parts of the fraction equal zero. This helps to obtain a function expression where the limit can be calculated accurately.

Indeterminate Forms

Indeterminate forms, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), appear when both the numerator and denominator of a fraction approach zero or infinity concurrently. These forms are "indeterminate" in the sense that they do not directly yield a specific limit and often require additional algebraic manipulation to solve.

In the given limit problem, as \(x\) nears 9, both the numerator \(\sqrt{x} - 3\) and the denominator \(x - 9\) were zero, resulting in the \(\frac{0}{0}\) form. Situations like this call for techniques such as factoring, multiplying by conjugates, or expanding to get past the zero values and identify the true limit.

After simplifying the equation using the difference of squares and cancelling common factors, the limit could be determined and the indeterminate form resolved. These strategies allow us to bypass the immediate complications of indeterminate forms and find valid, finite limits confidently.