Problem 6 - Thomas Calculus

Question

In Exercises \(1-6,\) use 1'Hopital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2 . $$ \lim _{x \rightarrow \infty} \frac{2 x^{2}+3 x}{x^{3}+x+1} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Recognize the Indeterminate Form

The given limit is \( \lim _{x \rightarrow \infty} \frac{2x^2+3x}{x^3+x+1} \). To see if 1'Hôpital's Rule applies, identify the form as \( \frac{\text{polynomial}}{\text{polynomial}} \). For large \( x \), both the numerator and the denominator approach infinity, suggesting the form \( \frac{\infty}{\infty} \), which is an indeterminate form.

Step 2: Applying 1'Hôpital's Rule

Since we have an indeterminate form \( \frac{\infty}{\infty} \), we apply 1'Hôpital's Rule, which requires taking the derivative of the numerator and the denominator.Compute the derivatives:- The derivative of the numerator \( 2x^2 + 3x \) is \( 4x + 3 \).- The derivative of the denominator \( x^3 + x + 1 \) is \( 3x^2 + 1 \).Now, apply 1'Hôpital's Rule:\[\lim _{x \rightarrow \infty} \frac{4x+3}{3x^2+1}\]

Step 3: Evaluate the New Limit

Evaluate the new limit \( \lim _{x \rightarrow \infty} \frac{4x+3}{3x^2+1} \). This still yields the \( \frac{\infty}{\infty} \) form, so we apply 1'Hôpital's Rule again.Compute the second derivatives:- The derivative of \( 4x + 3 \) is \( 4 \).- The derivative of \( 3x^2 + 1 \) is \( 6x \).Apply 1'Hôpital's Rule a second time:\[\lim _{x \rightarrow \infty} \frac{4}{6x}\]

Step 4: Simplify and Find the Limit

Now evaluate the simplified limit \( \lim _{x \rightarrow \infty} \frac{4}{6x} \). Since \( 6x \) grows without bound as \( x \to \infty \), the expression approaches 0:\[\lim _{x \rightarrow \infty} \frac{4}{6x} = 0\]

Step 5: Confirm the Result Using Algebraic Simplification

To confirm the result, perform algebraic simplification. Divide both the numerator and the denominator of the original function by \( x^3 \):\[\lim _{x \rightarrow \infty} \frac{\frac{2x^2}{x^3} + \frac{3x}{x^3}}{\frac{x^3}{x^3} + \frac{x}{x^3} + \frac{1}{x^3}} = \lim _{x \rightarrow \infty} \frac{\frac{2}{x} + \frac{3}{x^2}}{1 + \frac{1}{x^2} + \frac{1}{x^3}}\]As \( x \to \infty \), terms \( \frac{2}{x}, \frac{3}{x^2}, \frac{1}{x^2}, \text{and}\ \frac{1}{x^3} \) all approach 0, simplifying the limit to 0.

Key Concepts

Understanding Limits

Limits are fundamental in calculus, providing a way to describe the behavior of functions as they approach specific points or extend towards infinity. In this exercise, we evaluate the limit of a rational function, which involves both polynomials in the numerator and the denominator, as the variable approaches infinity.
To determine this limit, we initially face an indeterminate form \( \frac{\infty}{\infty} \). This signals that direct substitution results in an undefined expression, and we need alternative techniques such as L'Hôpital's Rule to find the limit.
The concept of limits helps us understand rates of change and the foundational elements of derivative calculations in calculus. By evaluating limits, we can predict function behavior at extreme values or points of discontinuity, providing a clearer picture of the function's overall behavior.

The Power of Calculus

Calculus is essential for solving complex problems involving continuous change. It offers tools like derivatives and integrals, which help in understanding functions at a deeper level. In this context, L'Hôpital's Rule allows us to address indeterminate forms in limits using derivatives.
By applying calculus concepts, specifically differentiation, we transform challenging or unsolvable limits into simpler forms that are easier to evaluate. Once differentiation is applied to both the numerator and the denominator through L'Hôpital's Rule, we simplify the expression, enabling clear evaluation of the limit.
This process demonstrates calculus's utility in providing structured methods to tackle difficult or abstract problems that often arise in mathematical analysis and applied sciences.

Handling Indeterminate Forms

Indeterminate forms arise in calculus when expressions yield ambiguous outcomes like \( \frac{\infty}{\infty} \), \( \frac{0}{0} \), or other non-definitive values. These forms require special methods to resolve, as they don't provide explicit answers on their own.
L'Hôpital's Rule becomes a crucial tool in managing indeterminate forms, by transforming the problematic expression using derivatives. In this exercise, we repeatedly applied the rule until achieving a deterministic result, which in this case is zero.
Understanding how to identify and effectively resolve indeterminate forms not only aids in limit evaluation but also enhances our problem-solving capabilities across calculus and beyond. It teaches us how to systematically dismantle complex expressions to reach clear, meaningful conclusions.