Problem 4 - Advanced Calculus

Question

Find \(\lim _{x \rightarrow 2} f(x)\) where \(f(x)=\frac{x^{3}+x^{2}-7 x+2}{2 x^{3}-5 x^{2}+6 x-8}\)

Step-by-Step Solution

Answer

The limit does not exist.

Step 1: Substitute \(x\) Value

Substitute \(x=2\) into the function, \(f(x)=\frac{x^{3}+x^{2}-7 x+2}{2x^{3}-5 x^{2}+6 x-8}\), to get \(f(2)=\frac{2^{3}+2^{2}-7 \cdot2+2}{2(2)^{3}-5 (2)^{2}+6 \cdot2-8}\).

Step 2: Simplify Numerator and Denominator

Simplify the numerator and denominator separately to get \(f(2)=\frac{8+4-14+2}{16-20+12-8}\).

Step 3: Compute the Limit

Calculate the numerator and denominator separately to get \(f(2) = \frac{0}{0}\). This is an indeterminate form, and thus, we cannot determine the limit. However, we can further evaluate it using L'Hopital's rule.

Step 4: Differentiate the Numerator and Denominator

Differentiate the numerator \(x^{3}+x^{2}-7x+2\) to get \(3x^{2}+2x-7\) and differentiate the denominator \(2x^{3}-5x^{2}+6x-8\) to get \(6x^{2}-10x+6\). Replace the original function with the ratio of these two new functions and evaluate the limit as \(x\) approaches 2.

Step 5: Evaluate the Limit

Substitute \(x=2\) into the new function. Hence the limit \(\lim_{x \rightarrow 2} f(x)=\frac{3(2)^{2}+2(2)-7}{6(2)^{2}-10(2)+6} = \frac{-1}{0}\). But this is undefined, so the limit does not exist.

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, you may encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. They do not directly give us the limit value.

Indeterminate forms occur because substituting the value into a function results in an invalid mathematical expression. For example, trying to compute \( f(2) \) using the original function \( \frac{x^3 + x^2 - 7x + 2}{2x^3 - 5x^2 + 6x - 8} \) directly leads to \( \frac{0}{0} \).

Recognizing indeterminate forms is crucial. It signals us to use other methods, such as L'Hôpital's Rule, to evaluate limits accurately and avoid incorrect assumptions.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful and helpful tool in calculus for resolving indeterminate forms. Named after the French mathematician Guillaume de l'Hôpital, this rule provides a systematic way to compute limits that would otherwise result in undefined expressions like \( \frac{0}{0} \).

The rule states that if you have an indeterminate form, you can take the derivative of the numerator and the denominator separately, then calculate the limit of the new function. It simplifies the process and often leads to a well-defined limit.

For example, in our problem, we replace the original function by differentiating both the top function, \( x^3 + x^2 - 7x + 2 \), and the bottom function, \( 2x^3 - 5x^2 + 6x - 8 \). By computing their derivatives, we obtain \( 3x^2 + 2x - 7 \) and \( 6x^2 - 10x + 6 \) respectively.

Then, we assess the limit of this new function. This technique helps in tackling otherwise troublesome limits in a structured and computational manner.

Differentiation

Differentiation is a fundamental concept in calculus, essential for implementing L'Hôpital's Rule as well as for understanding how functions behave.

It involves calculating the derivative of a function, which represents the function's rate of change. The derivative can be thought of as the slope of the tangent line to the curve at a given point.

In our limit problem, to apply L'Hôpital's Rule, we had to differentiate both the numerator and the denominator of the initial function. For \( x^3 + x^2 - 7x + 2 \), the derivative is \( 3x^2 + 2x - 7 \). Similarly, for \( 2x^3 - 5x^2 + 6x - 8 \), the derivative is \( 6x^2 - 10x + 6 \).

By understanding and using differentiation, you can transform complex expressions into simpler ones, making it easier to solve limits and uncover a function's behavior near specific points.