Problem 71 - Precalculus

Question

Evaluate the limits using limit properties. If a limit does not exist, state why. $$\lim _{x \rightarrow-7} \frac{\sqrt{x^{2}+14 x+49}}{x^{2}+8 x+7}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{6} \).

Step 1: Mention Substitution

To solve the limit problem, we first try direct substitution of the value \( x = -7 \). By substituting \( -7 \) directly into the expression \( \sqrt{x^2 + 14x + 49} \):\[\sqrt{(-7)^2 + 14(-7) + 49} = \sqrt{49 - 98 + 49} = \sqrt{0} = 0 \]The numerator simplifies to 0. Now, for the denominator \( x^2 + 8x + 7 \):\[(-7)^2 + 8(-7) + 7 = 49 - 56 + 7 = 0\]The denominator also simplifies to 0, which suggests indeterminate form \( \frac{0}{0} \), requiring further analysis.

Step 2: Factorization of Expression

Given the indeterminate form \( \frac{0}{0} \), we attempt to simplify by factoring both the numerator and the denominator.The expression under the square root is:\[ x^2 + 14x + 49 = (x + 7)^2 \]Thus, \( \sqrt{x^2 + 14x + 49} = |x + 7| \).The denominator can be factored similarly:\[ x^2 + 8x + 7 = (x + 7)(x + 1) \]The expression now becomes:\[ \frac{|x + 7|}{(x + 7)(x + 1)} \]

Step 3: Simplify the Expression

Since \( x \) approaches \( -7 \), we have \( |x + 7| = -(x + 7) \) because the expression within the modulus is negative.Simplify the expression:\[ \frac{|x+7|}{(x+7)(x+1)} = \frac{-(x+7)}{(x+7)(x+1)} \]We can cancel the \( (x+7) \) terms in the numerator and denominator:\[ \frac{-1}{x+1} \]

Step 4: Evaluate the Limit

Now that the indeterminate form is resolved, evaluate the limit by plugging in \( x = -7 \):\[ \lim_{x \to -7} \frac{-1}{x+1} = \frac{-1}{-7+1} = \frac{-1}{-6} = \frac{1}{6} \]

Key Concepts

Indeterminate Forms

When we work with limits, sometimes we encounter situations like the dreaded "indeterminate forms." This happens when direct substitution into an expression leads to an undefined result such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are not definitive solutions but rather signals that we need a different approach to evaluate the limit properly.
In our problem, substituting \( x = -7 \) directly into both the numerator and the denominator results in zero. This \( \frac{0}{0} \) form is known as indeterminate. It tells us that more work is needed to figure out the actual value of the limit.

Factoring

Factoring is a powerful tool in handling indeterminate forms, allowing us to simplify expressions by cancelling out common terms. This makes it easier to assess the limit accurately.
Let's start with our numerator, \( \sqrt{x^2 + 14x + 49} \). This can be rewritten as \( |x + 7| \). Similarly, our denominator \( x^2 + 8x + 7 \) factors to \((x + 7)(x + 1)\).

• Factoring transforms the expression into a simpler form: \( \frac{|x + 7|}{(x + 7)(x + 1)} \).
• Visualizing it in this way helps identify the common term \( (x+7) \), which we can then cancel out.

Ultimately, removing the common factor resolves the indeterminacy, simplifying the expression.

Limit Properties

Limits follow several important properties that help simplify their evaluation. These properties allow us to handle expressions logically and methodically.
For instance, when dealing with fractions, we can exploit properties like:

• Quotient Law: This lets us separate the limit of a fraction into the limits of the numerator and denominator (provided no indeterminate form arises).
• Simplification: Cancelling common factors where possible helps break down complex expressions.

In our example, simplifying by cancellation and then applying the limit helps us avoid direct indeterminate forms and find a more straightforward expression to evaluate.

Direct Substitution

Direct substitution is the first step when evaluating limits, as it usually offers a quick and straightforward way to find the solution. Here, you simply replace the variable with the approach value.
However, as we've seen, direct substitution can lead to indeterminate forms like \( \frac{0}{0} \).

• If substituting results in a determinable value, then there’s no need for additional methods.
• When indeterminate, this step informs us another approach is necessary, such as factoring or using limit laws.

After resolving the indeterminacy in our example, we returned to direct substitution for a clean result, showing why this is always a great first try.