Problem 4 - Active Calculus

Question

Evaluate the limit $$\lim _{x \rightarrow-6} \frac{x^{2}-36}{x+6}$$ If the limit does not exist enter DNE. Limit =\(\square\)

Step-by-Step Solution

Answer

The limit is \(-12\).

Step 1 - Identify the form of the limit

First, observe the limit \(\lim _{x \rightarrow-6} \frac{x^{2}-36}{x+6}\). Substitute \(x = -6\) into the expression. \(f(x) = \frac{(-6)^{2}-36}{-6+6} = \frac{36-36}{0} = \frac{0}{0}\). This is an indeterminate form of \(\frac{0}{0}\).

Step 2 - Factor the numerator

Factor the numerator to find a simpler form: \[x^2 - 36 = (x + 6)(x - 6).\]

Step 3 - Simplify the expression

Rewrite the original limit with the factored numerator: \[\lim _{x \rightarrow-6} \frac{(x + 6)(x - 6)}{x + 6}.\] Simplify by canceling the common factor \(x + 6\): \[\lim _{x \rightarrow-6} (x - 6).\]

Step 4 - Evaluate the limit

Now evaluate the simplified limit by direct substitution of \(x = -6\): \(\lim _{x \rightarrow-6} (x - 6) = -6 - 6 = -12.\)

Key Concepts

limits and continuity

In calculus, understanding limits is fundamental. When we talk about the limit of a function as it approaches a particular point, we are interested in what value the function gets closer to.
For the given exercise, we wanted to know what happens to the expression as x approaches -6.
Limits are essential for defining continuity in functions. A function is continuous at a point if the limit coincides with the function’s value there, and the function is defined at that point.
In our exercise, recognizing limits helped determine our solution step by step.

indeterminate forms

Indeterminate forms occur when substituting a value into a limit produces an expression that isn't immediately solvable. Generally, these forms include \( \frac{0}{0} \), \( \frac{\text{∞}}{\text{∞}} \), and \( 0 \times \text{∞} \). Steps must be taken to simplify or manipulate the expression to resolve the limit.
In our example, substituting -6 into the given limit \( \frac {x^2 - 36}{x+6} \) resulted in \( \frac{0}{0} \), an indeterminate form.
Identifying this indeterminate form was crucial in recognizing that further algebraic manipulation was necessary (like factoring) to progress towards finding the limit.

factoring in calculus

Factoring is a powerful tool in calculus, especially when dealing with limits involving indeterminate forms. By rewriting the mathematical expression in its factored form, often, we can simplify the problem and cancel out common factors.
In our original exercise, the expression \( \frac{x^2 - 36}{x+6} \) was simplified by factoring the numerator: \[ x^2 - 36 = (x + 6)(x - 6). \] Once in this simpler form, we can cancel common factors to find the limit without dealing with the indeterminate form. This technique transformed the complex limit problem into a simpler one, leading us to a straightforward solution.

direct substitution in limits

Direct substitution is the method of finding limits by simply substituting the approaching value into the function. However, this method only works if the function is continuous or if the substitution doesn't lead to indeterminate forms.
Before applying this technique, check if the function yields a determinate value. If substituting creates an indeterminate form, like in our exercise \( \frac{x^2 - 36}{x+6} \), then further simplification is needed.
After factoring and canceling out common terms, direct substitution allowed us to solve the limit: \[ \text{The limit} = \ \frac{(x+6)(x-6)}$(x+6) \text{... finally results in } (x-6). \ \text {We then evaluate: } \ \text {limit as x \rightarrow -6} = (-6 - 6 = -12). \] This confirms our solution, illustrating the power of direct substitution when used correctly.