Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow-2} \frac{x^{2}+6 x+8}{x^{2}-3 x-10} $$

Step-by-Step Solution

Answer

The limit is \(-\frac{2}{7}\).

Step 1: Substitute the Value

Let's substitute \( x = -2 \) into the expression to see if we have an indeterminate form. We compute the numerator and the denominator separately.\[\text{Numerator} = (-2)^2 + 6(-2) + 8 = 4 - 12 + 8 = 0 \] \[\text{Denominator} = (-2)^2 - 3(-2) - 10 = 4 + 6 - 10 = 0 \] So, the expression \( \frac{x^2 + 6x + 8}{x^2 - 3x - 10} \) is in the indeterminate form \( \frac{0}{0} \) when \( x \rightarrow -2 \).

Step 2: Apply l'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply l'Hôpital's Rule. This rule states that:\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\] provided the limit on the right exists. First, we calculate the derivatives:\[f'(x) = \frac{d}{dx}(x^2 + 6x + 8) = 2x + 6\]\[g'(x) = \frac{d}{dx}(x^2 - 3x - 10) = 2x - 3\]

Step 3: Evaluate the Limit of the Derivatives

Now, substitute \( x = -2 \) into the derivatives:\[\lim_{x \to -2} \frac{2x + 6}{2x - 3} = \frac{2(-2) + 6}{2(-2) - 3} = \frac{-4 + 6}{-4 - 3} = \frac{2}{-7} = -\frac{2}{7} \]The limit evaluates to \(-\frac{2}{7}\).

Key Concepts

Indeterminate Forms

When solving limits in calculus, you may encounter expressions that initially don't have a clear limit. These are called **indeterminate forms**. An indeterminate form looks like

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(0 \times \infty\)
• \(\infty - \infty\)
• \(0^0\)
• \(1^\infty\)
• \(\infty^0\)

These forms arise when substituting a value into a limit results in uncertainty. For instance, when evaluating the limit \(\lim_{x \to -2} \frac{x^2 + 6x + 8}{x^2 - 3x - 10}\), substituting \(x = -2\) yields \(\frac{0}{0}\), an indeterminate form. Here, the expression doesn’t immediately tell us what the limit is, which is why we use additional techniques like l'Hôpital's Rule.

Derivatives

Derivatives are a core concept in calculus representing the rate of change of a function. To apply l'Hôpital's Rule effectively, you need to compute derivatives of both the numerator and denominator functions.In our example, the expression \(\frac{x^2 + 6x + 8}{x^2 - 3x - 10}\) requires finding the derivatives:

• The derivative of \(f(x) = x^2 + 6x + 8\) is \(f'(x) = 2x + 6\).
• The derivative of \(g(x) = x^2 - 3x - 10\) is \(g'(x) = 2x - 3\).

These derivatives give us new functions to substitute and assess the limit again, especially when dealing with an indeterminate form such as \(\frac{0}{0}\). Taking derivatives simplifies the mathematical expression, making it easier to evaluate.

Limits

In calculus, a limit describes the value that a function approaches as the input approaches some point. Limits are vital for understanding the behavior of functions at specific points, particularly when the function isn’t explicitly defined at that point.When encountering an indeterminate form like \(\frac{0}{0}\), limits let us use calculus tools, such as l'Hôpital's Rule, to find a defined value. In the example \(\lim_{x \to -2} \frac{x^2 + 6x + 8}{x^2 - 3x - 10}\), after differentiating, the limit becomes\[\lim_{x \to -2} \frac{2x + 6}{2x - 3}\]Substituting \(x = -2\) now gives a clear outcome: \(-\frac{2}{7}\), indicating the limit exists and is determined. Limits allow us to make sense of the indeterminate forms and continue our analysis in calculus.