Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 1^{-}} \frac{x^{2}-2 x+2}{x^{2}-1} $$

Short answer

The limit is negative infinity.

Step-by-step solution

Substitute the value into the limit

First, substitute the value of 1 into the given expression to check if it forms an indeterminate form. Substitute x = 1 into \( \frac{x^{2}-2x+2}{x^{2}-1} \). This gives \( \frac{1^2-2 \times 1 + 2}{1^2-1} = \frac{1-2+2}{1-1} = \frac{1}{0} \).

Determine the form of the expression

Since substituting x = 1 yields \( \frac{1}{0} \), which is not an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), l'Hôpital's Rule cannot be applied directly.

Analyze the limit behavior

Observe the behavior of the expression as x approaches 1 from the left. As \( x \to 1^- \), the numerator approaches 1 and the denominator approaches 0 from the negative side.

Determine the sign of the limit

Since the numerator approaches a positive value and the denominator approaches zero from the negative side, the overall fraction \( \frac{x^2 - 2x + 2}{x^2 - 1} \) tends to negative infinity.

Key concepts

Limit Evaluation

When tackling a limit problem, the main goal is to determine the value that a function approaches as the variable gets arbitrarily close to a specified point. In this exercise, you are given the limit \[\lim _{x \rightarrow 1^{-}} \frac{x^{2}-2 x+2}{x^{2}-1}\] and asked to evaluate it. To start, substitute the value of the variable, in this case, 1, into the function.
Upon substitution, if you end up with a form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you have an indeterminate form, which requires special techniques like l'Hôpital's Rule to resolve. However, here, substituting \(x=1\) yields \(\frac{1}{0}\), indicating a path towards a vertical asymptote rather than an indeterminate form.
Next, check the behavior of the function near this point to understand how the numerator and denominator change, especially if they approach zero or infinity. This helps in evaluating the trend or the direction in which the function heads as it nears the point from the left.

Indeterminate Forms

Indeterminate forms are expressions that do not readily point to an obvious limit when directly substituted. These forms require special techniques for evaluation. The most common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), 0 multiplied by infinity, infinity minus infinity, and others like these.
These forms appear because direct substitution leads either to a division by zero or involves infinity in some ambiguous way. To resolve such cases, techniques such as factoring, conjugates, or specific rules like l'Hôpital's can be used.
In the exercise provided, although initially substituting \(x = 1\) in the function \(\frac{x^{2}-2x+2}{x^{2}-1}\) returns \(\frac{1}{0}\), it indicates a potential vertical asymptote rather than a classic indeterminate form. Therefore, l'Hôpital's Rule does not apply here since this rule is mainly for forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Instead, observing the sign and the behavior around this point helps determine the correct limit.

Calculus Techniques

In calculus, deciding the best strategy to evaluate a limit usually depends on the form of the expression you are dealing with. For evaluating limits, some popular techniques include:

• Algebraic Manipulation: Simplifying equations by combining terms or factoring them.
• Conjugates: Particularly useful for functions involving roots, where multiplying by a conjugate can simplify the expression.
• l'Hôpital's Rule: Used for limits in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Sign Analysis: Particularly useful when approaching points of discontinuity, where simply observing the sign of the expression can help deduce the limit behavior.

In the given problem, the critical technique was observing the function's behavior as \(x\) approached 1 from the left. By noting that the denominator approaches zero from the negative side, and the numerator tends towards a positive value, we can infer that the whole expression trends towards negative infinity. This is a valuable technique when direct evaluation doesn't yield a straightforward answer due to a discontinuous point or a vertical asymptote.