Question

Evaluate the given limit. $$ \lim _{x \rightarrow 2} \frac{x^{2}-10 x+16}{x^{2}-x-2} $$

Short answer

The limit is -2.

Step-by-step solution

Substitute x with the Limit Value

First, substitute the value that x approaches, which is 2, directly into the expression \( \frac{x^{2}-10x+16}{x^{2}-x-2} \). This will help to check if there might be any indeterminate form or if the expression simplifies easily.Substituting x = 2 gives us:\[ \frac{2^{2} - 10 \times 2 + 16}{2^{2} - 2 - 2} = \frac{4 - 20 + 16}{4 - 2 - 2} = \frac{0}{0} \]This result is an indeterminate form, so we need to factor the expression further.

Factor the Numerator and Denominator

Since we found an indeterminate form, we need to factor both the numerator and the denominator to try to cancel out common factors.Factor the numerator: \[ x^2 - 10x + 16 = (x-2)(x-8) \]Factor the denominator:\[ x^2 - x - 2 = (x-2)(x+1) \].

Simplify the Expression by Canceling Common Factors

Now that both the numerator and the denominator are factored, identify and cancel out any common factors:\[ \frac{(x-2)(x-8)}{(x-2)(x+1)} \]The factor \((x-2)\) appears in both the numerator and the denominator, so it can be canceled:\[ \frac{x-8}{x+1} \]

Evaluate the Limit of the Simplified Expression

After canceling the common factor, substitute x = 2 into the simplified expression to find the limit:\( \lim_{x \rightarrow 2} \frac{x-8}{x+1} \) becomes:\[ \frac{2-8}{2+1} = \frac{-6}{3} = -2 \].

Conclusion

Thus, the limit of the original expression as x approaches 2 is \(-2\). The indeterminate form was resolved by factoring and simplifying, allowing us to evaluate the limit successfully.

Key concepts

Indeterminate Forms

In calculus, encountering indeterminate forms is a common challenge when evaluating limits. An indeterminate form arises when substituting a value into an expression results in an undefined state like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms reveal that more work is needed to determine the actual limit.

When you first substitute \( x = 2 \) into \( \frac{x^{2}-10x+16}{x^{2}-x-2} \), you get the indeterminate form \( \frac{0}{0} \). This indicates there's a chance that the expression can be further simplified or modified to allow for the calculation of the limit. To resolve these, techniques like factoring or using L'Hôpital's rule come into play, though in this case, we chose factoring. Understanding indeterminate forms makes it easier to decide how to proceed with solving limits.

Factoring Expressions

Factoring is a powerful algebraic tool that can transform complex expressions into simpler ones. It involves breaking down an expression into products of simpler factors that can be more easily managed. When dealing with limits, factoring often helps in canceling out terms that cause indeterminacy.

In our example, we factored both the numerator, \( x^2 - 10x + 16 \), and the denominator, \( x^2 - x - 2 \). This process yielded the factors \((x-2)(x-8)\) and \((x-2)(x+1)\) respectively.

• Numerator becomes \((x-2)(x-8)\)
• Denominator becomes \((x-2)(x+1)\)

Factoring reveals common terms, like \((x-2)\), that simplify the expression when canceled. Mastery of factoring is crucial in making complex expressions more approachable.

Simplifying Expressions

Simplifying expressions is a key step in solving limit problems because it often uncovers the real value of the limit after eliminating indeterminate forms. After factoring, simplification involves identifying and canceling common factors. This step is essential for cleaning up the equation to make substitution directly possible.

In the example, once we factored and saw the common factor \( (x-2) \) in both the numerator and the denominator, we canceled it out, leaving us with \( \frac{x-8}{x+1} \). This process transforms the original problematic limit into something solvable.

• Identify common terms.
• Cancel the terms to eliminate indeterminacy.

Simplification is an essential technique for transforming math expressions into their most manageable forms.

Evaluating Limits

Evaluating limits is the final goal when solving problems involving limits. By using techniques like substitution, factoring, and simplification, you make it possible to find the actual limit value accurately. Once you've simplified the expression, as in \( \frac{x-8}{x+1} \), you substitute the limit point \( x=2 \) into this expression.

For this example, the substitution process becomes straightforward: \[ \lim_{x \rightarrow 2} \frac{x-8}{x+1} = \frac{2-8}{2+1} = \frac{-6}{3} = -2 \]Evaluating limits requires careful transformations to avoid indeterminate forms and reach a clear value. With practice, recognizing patterns and choosing appropriate methods become natural, making limit evaluations second nature.