Problem 33 - Precalculus

Question

Evaluate the following limits by rewriting the given expression as needed. $$\lim _{x \rightarrow-3} \frac{\sqrt{x+7}-2}{x+3}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{4} \).

Step 1: Identify Direct Substitution

First, check by direct substitution to see if the given limit can be evaluated directly. Substitute \( x = -3 \) into the function: \( \frac{\sqrt{-3+7}-2}{-3+3} \). This results in \( \frac{\sqrt{4}-2}{0} = \frac{2-2}{0} = \frac{0}{0} \). The \( \frac{0}{0} \) form indicates an indeterminate form. We need to manipulate the expression further.

Step 2: Rationalize the Numerator

To eliminate the indeterminate form, rationalize the numerator of the expression by multiplying the numerator and the denominator by the conjugate of the numerator, \( \sqrt{x+7} + 2 \). This gives: \[ \frac{\sqrt{x+7}-2}{x+3} \times \frac{\sqrt{x+7}+2}{\sqrt{x+7}+2} = \frac{ (\sqrt{x+7})^2 - 2^2}{(x+3)(\sqrt{x+7}+2)}. \]

Step 3: Simplify the Expression

Simplify the numerator by using the difference of squares: \((\sqrt{x+7})^2 - 2^2 = x+7 - 4 = x+3\). The expression now becomes: \[ \frac{x+3}{(x+3)(\sqrt{x+7}+2)}. \]

Step 4: Cancel Common Factors

Cancel the common factor of \( x+3 \) from the numerator and denominator: \[ \frac{x+3}{(x+3)(\sqrt{x+7}+2)} = \frac{1}{\sqrt{x+7} + 2}. \]

Step 5: Substitute and Evaluate the Limit

Now, substitute \( x = -3 \) into the expression to evaluate the limit: \[ \frac{1}{\sqrt{-3+7} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2+2} = \frac{1}{4}. \] This is the value of the limit.

Key Concepts

Direct Substitution

Direct substitution is the first strategy to try when evaluating a limit. It's simple: you plug the number the variable is approaching directly into the function. For the problem we have, substitute \( x = -3 \) into the expression \( \frac{\sqrt{x+7}-2}{x+3} \).
When we do this, the expression becomes \( \frac{\sqrt{-3+7}-2}{-3+3} \). This equals \( \frac{\sqrt{4}-2}{0} = \frac{2-2}{0} = \frac{0}{0} \).
This \( \frac{0}{0} \) is what we call an **indeterminate form**, indicating we need another approach. However, this step is crucial as it often quickly confirms if a limit needs further work.

Rationalizing the Numerator

Rationalizing the numerator is a clever trick used in calculus to deal with square roots in limits. Essentially, you multiply the expression by a form of 1 to eliminate the square root. In our expression, this means multiplying the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x+7} + 2 \).
By multiplying, we re-formulate the expression:

• Numerator: \( \sqrt{x+7} - 2 \times \sqrt{x+7} + 2 = (\sqrt{x+7})^2 - 2^2 \).
• Denominator: \( (x+3)(\sqrt{x+7} + 2) \).

The rationale is to use the identity \( a^2 - b^2 = (a-b)(a+b) \) to simplify the square root, a technique further explained in the next section.

Difference of Squares

The difference of squares is a nifty algebraic identity \( a^2 - b^2 = (a-b)(a+b) \) that helps simplify expressions involving squares. In our limit exercise, after rationalizing the numerator, we apply this identity.
The expression \( (\sqrt{x+7})^2 - 2^2 \) simplifies to \( x+7 - 4 = x+3 \). This step results in:

• Numerator becomes: \( x+3 \).
• Rewritten expression: \( \frac{x+3}{(x+3)(\sqrt{x+7}+2)} \).

With the common factor \( x+3 \) present in both the numerator and denominator, we can now cancel these terms, simplifying further so direct substitution can yield a meaningful result.

Indeterminate Form

Indeterminate forms like \( \frac{0}{0} \) often signal that a limit needs special techniques beyond direct substitution. Seeing this form means that neither the numerator nor the denominator contributes a clear value when zero, which can lead to apparent contradictions.
Common approaches for resolving indeterminate forms include:

• Rationalizing, as discussed here.
• Factoring.
• L'Hôpital's rule, for other more complex functions.

In this problem, after rationalizing and simplifying, the expression becomes \( \frac{1}{\sqrt{x+7} + 2} \). Now substitute \( x = -3 \) again to get \( \frac{1}{4} \).
Thus, effectively addressing the indeterminate form through algebraic manipulation leads to the expression's limit value.