Question

find the indicated limit or state that it does not exist. In many cases, you will want to do some algebra before trying to evaluate the limit. $$ \lim _{x \rightarrow-1} \frac{x^{2}-2 x-3}{x+1} $$

Short answer

The limit is -4.

Step-by-step solution

Check Direct Substitution

First, we try direct substitution of \(x = -1\) into the function \(f(x) = \frac{x^2 - 2x - 3}{x+1}\).Calculate:\[ \lim _{x \rightarrow-1} \frac{(-1)^{2} - 2(-1) - 3}{-1+1} = \frac{1 + 2 - 3}{0} = \frac{0}{0} \]The result is an indeterminate form \(\frac{0}{0}\). This suggests further algebraic manipulation is required.

Factor the Numerator

To resolve the indeterminate form, we can factor the numerator. Observe the expression:\[ x^2 - 2x - 3 \]This can be factored as:\[ (x - 3)(x + 1) \]

Simplify the Expression

Now, substitute the factored form of the numerator into the original expression:\[ \frac{(x - 3)(x + 1)}{x + 1} \]Cancel out the common term \((x + 1)\) from the numerator and the denominator (except when \(x = -1\), as it was causing the zero denominator):\[ x - 3 \]

Evaluate the Simplified Limit

After simplification, evaluate the limit of the remaining expression \(x - 3\) as \(x\) approaches \(-1\):\[ \lim _{x \rightarrow-1} (x - 3) = -1 - 3 = -4 \]

Key concepts

Indeterminate Forms

When you're working with limits in calculus, you may encounter a scenario called an "indeterminate form." This typically happens when you directly substitute a value into a limit expression and get something unexpected, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms are called indeterminate because they don't lead to any obvious conclusion about the limit's value from direct substitution alone.

In our example, directly substituting \(x = -1\) resulted in \(\frac{0}{0}\). This is a classic indication of an indeterminate form. It signals that you might need to use algebraic manipulation or other strategies to properly evaluate the limit. Don't worry, this situation doesn't mean the limit inherently doesn't exist; it simply means further analysis is needed to find a more definitive form.

Next, we'll explore how to tackle these forms using algebraic techniques! 👍

Algebraic Manipulation

After identifying an indeterminate form, such as \(\frac{0}{0}\), one common approach is to use algebraic manipulation to simplify the expression. By simplification, we mean rewriting the expression in a form that allows for easy evaluation of the limit.

Let's use the example \( \frac{x^2 - 2x - 3}{x+1} \). The key step is to manipulate the expression so that it becomes more manageable. Here, we factor the numerator, \(x^2 - 2x - 3\), to get \((x - 3)(x + 1)\). This makes it evident that the numerator and denominator share a common factor \((x + 1)\).

By understanding how to manipulate algebraic structures like this, you can turn complex problems into simpler ones. The next step is to identify these common factors, which is explained below.

Canceling Common Factors

After factoring the numerator, you may discover that both the numerator and the denominator have a common factor. In our case, \((x + 1)\) appears in both places. Once we have expressed both parts of our fraction as products, we can "cancel" these common factors.

Here's how it works: when you cancel out \( (x + 1) \) from the expression \( \frac{(x - 3)(x + 1)}{x + 1} \), you're left with \(x - 3\), provided \(x eq -1\). While this action eliminates the source of the indeterminate form, ensure this cancellation step does not violate any conditions of the function, such as division by zero.

Now that the troublesome term is gone, what remains is a much simpler expression \( x - 3 \), which can be evaluated at \( x = -1 \). The limit is now straightforward as \( \lim_{x \to -1} (x - 3) = -4 \). This process of canceling out common factors is crucial to turning potentially complicated calculus problems into simpler ones! 🎉