Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 5} \frac{2 x^{2}-11 x+5}{x-5}=9 $$

Step-by-Step Solution

Answer

The limit is 9.

Step 1: Recognize Indeterminate Form

When trying to evaluate the limit \( \lim _{x \rightarrow 5} \frac{2x^2 - 11x + 5}{x - 5} \), direct substitution results in \( \frac{0}{0} \), which is an indeterminate form. This requires further analysis.

Step 2: Factor the Numerator

To resolve the indeterminate form, factor the numerator \( 2x^2 - 11x + 5 \). Look for two numbers that multiply to \( 2 \times 5 = 10 \) and add to \(-11\). These numbers are -1 and -10. Thus, the numerator can be rewritten and factored as \((2x - 1)(x - 5)\).

Step 3: Simplify the Expression

The expression now is \( \frac{(2x - 1)(x - 5)}{x - 5} \). Cancel the common factor \(x - 5\) in the numerator and the denominator. Hence, it simplifies to \(2x - 1\).

Step 4: Substitute the Limit Value

Substitute \( x = 5 \) into the simplified expression \( 2x - 1 \). Calculate \( 2 \times 5 - 1 = 10 - 1 = 9 \).

Step 5: Conclusion

Since the simplified expression evaluated at \( x = 5 \) equals 9, the limit is: \( \lim _{x \rightarrow 5} \frac{2x^2 - 11x + 5}{x - 5} = 9 \).

Key Concepts

Indeterminate Forms

Indeterminate forms often arise in calculus when we are trying to evaluate limits. An indeterminate form means that the expression doesn't initially produce a clear numerical answer when substituting a value. In the example given, when attempting to find \( \lim _{x \rightarrow 5} \frac{2x^2 - 11x + 5}{x - 5} \), direct substitution results in \( \frac{0}{0} \). This is a classic indeterminate form, implying that further steps are needed.

• Indeterminate forms like \( \frac{0}{0} \) can suggest that further algebraic manipulation is necessary.
• Other common indeterminate forms include \( \frac{\infty}{\infty}, 0 \times \infty, \infty - \infty, 0^0, \infty^0 \) and \( 1^\infty \).

Understanding indeterminate forms helps in knowing when additional analysis is essential to find the limit.

Factoring Polynomials

Factoring polynomials is a useful technique when dealing with indeterminate forms. It involves expressing a polynomial as a product of its factors. In our example, the polynomial in the numerator \(2x^2 - 11x + 5 \) needs factoring to resolve the \(\frac{0}{0}\) form.

• Start by identifying numbers that multiply to the product of the leading coefficient and the constant term. Here, it’s \(2 \times 5 = 10\).
• Next, find pair of numbers that multiply to \(10\) and add to \(-11\). These numbers are \(-1\) and \(-10\).
• Rewrite the middle term using these numbers: \(2x^2 - 10x - x + 5\), then factor by grouping: \[(2x - 1)(x - 5)\].

Factoring helps to simplify the expression, making it easier to work with further on.

Simplification Techniques

Once numerators and denominators are factored, simplifying the expression is the next step. By cancelling out common factors, we can transform complex forms into simpler expressions. In our example, \[ \frac{(2x-1)(x-5)}{x-5} \] we see \(x - 5\) is a common factor and can be cancelled from both numerator and denominator.

• Cancelling involves ensuring the factor to be cancelled does not cause division by zero.
• After cancellation, the remaining expression \(2x - 1\) is simpler.
• Important: Cancel only terms, not individual variables or numbers in expressions.

Simplification reduces complexity, allowing ease of further evaluation and understanding.

Substitution Method in Limits

The substitution method in limits is essential after simplifying expressions, especially when it's initially in an indeterminate form. This involves substituting the limit value directly into the simplified expression.Once we simplify to \(2x - 1\), we substitute \(x = 5\).

• Simply replace \(x\) in the simplified expression with \(5\): \[2 \times 5 - 1\].
• Calculate: \[ 10 - 1 = 9\].

Substitution is a powerful technique to directly evaluate the limit, providing the final solution after other techniques have refined the expression.