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. $$ \lim _{x \rightarrow-1} \frac{x^{2}+x}{x^{2}+1} $$

Short answer

The limit is 0.

Step-by-step solution

Determine the Direct Substitution

First, we substitute the value of \( x = -1 \) directly into the expression \( \frac{x^2 + x}{x^2 + 1} \). This results in \( \frac{(-1)^2 + (-1)}{(-1)^2 + 1} \). Simplifying the expression in the numerator gives \( 1 - 1 = 0 \) and in the denominator gives \( 1 + 1 = 2 \). Hence, the expression simplifies to \( \frac{0}{2} = 0 \).

Confirm Simplification

Since the substitution resulted in \( \frac{0}{2} \), which is a well-defined number, we confirm that no further algebraic manipulation is necessary and the limit can be calculated using direct substitution.

State the Limit

Since the evaluation from the second step confirms that the fraction \( \frac{0}{2} \) is properly defined and equals zero, the limit as \( x \rightarrow -1 \) is simply 0.

Key concepts

Understanding Direct Substitution

Direct substitution is often the first and simplest method attempted when evaluating limits. With this approach, you replace the variable in the function with the value it approaches. Here, the limit of the function \( \frac{x^2 + x}{x^2 + 1} \) as \( x \) approaches \(-1\) is calculated by substituting \(-1\) directly into the expression.

Let's break it down:

• Substitute \(-1\) for \(x\) in the expression.
• Calculate the numerator: \((-1)^2 + (-1) = 1 - 1 = 0\).
• Calculate the denominator: \((-1)^2 + 1 = 1 + 1 = 2\).

After substitution, you find \( \frac{0}{2} \). Since division by a non-zero number is defined, the result is simply \(0\). This tells us that direct substitution works perfectly here because the limit is a real number. If the denominator had been zero, further techniques would have been needed.

Mastering Algebraic Manipulation

Not all limit problems are straightforward enough for direct substitution. Sometimes, the expression may result in an undefined form like \(\frac{0}{0}\). In such cases, algebraic manipulation steps in to save the day.

For instance:

• Factor expressions to simplify complex fractions.
• Cancel common terms in the numerator and denominator.
• Rationalize the numerator or denominator, if needed.

In our problem, however, the substitution \(x = -1\) resulted in \(\frac{0}{2}\) instead of an undefined fraction. The absence of any indeterminate forms means we did not require any algebraic manipulation beyond initial simplification. Yet, always be prepared to use algebraic tools if direct substitution reveals an indeterminate form.

Navigating Limit Evaluation

Evaluating limits is a fundamental skill in calculus, which helps in understanding the behavior of functions as they approach particular points.

Here's how to tackle limit evaluation:

• First, try direct substitution. In many cases, like ours, it will provide the answer directly.
• If substitution fails, regress to algebraic manipulation to simplify the expression.
• In rare instances, advanced methods like L'Hôpital's Rule or series expansions might be necessary.

In our example, following direct substitution, we determined the limit is \(0\). Given the simplicity of the expression after substitution, no further steps were necessary. Evaluating limits allows you to predict how functions behave, which is crucial in calculus and in many applications across different fields of science and engineering.