Problem 35 - Thomas Calculus

Question

0/0 Form Estimate the value of $$\lim _{x \rightarrow 1} \frac{2 x^{2}-(3 x+1) \sqrt{x}+2}{x-1}$$ by graphing. Then confirm your estimate with l'Hopital's Rule.

Step-by-Step Solution

Answer

The limit is \(-1\).

Step 1: Analyze the Expression for Limits

We are given the limit \[ \lim _{x \rightarrow 1} \frac{2x^{2}-(3x+1)\sqrt{x}+2}{x-1} \]. Initially, by directly substituting \(x = 1\) in the function, we check the form and get \( \frac{0}{0} \). This indicates an indeterminate form, thus requiring other methods like graphing or l'Hopital's Rule to evaluate.

Step 2: Graph the Function

Graph the function \( f(x) = \frac{2x^{2}-(3x+1)\sqrt{x}+2}{x-1} \) using graphing tools or software. Observe the behavior of the function as \( x \) approaches 1. The graph will give a visual indication of the limit value by showing where the y-values converge when \( x \) is near 1.

Step 3: Apply l'Hopital's Rule

Since the original expression is in the 0/0 indeterminate form, apply l'Hopital's Rule which recommends taking derivatives of the numerator and the denominator. Differentiate the numerator: \((2x^2 - (3x+1)\sqrt{x}+2)' = 4x - (3\sqrt{x} + \frac{3x+1}{2\sqrt{x}})\). Differentiate the denominator: \((x-1)' = 1\).

Step 4: Evaluate the Derivatives at the Limit Point

Evaluate the limit using the derivatives obtained: \[ \lim _{x \rightarrow 1} \frac{4x - (3\sqrt{x} + \frac{3x+1}{2\sqrt{x}})}{1} \]. Simplify and substitute \(x=1\), yielding \( 4 - (3 + \frac{4}{2}) = 4 - 5 = -1 \). So, the value of the limit is \(-1\).

Key Concepts

Understanding Indeterminate Forms

In calculus, an indeterminate form arises when applying direct substitution in a limit results in an undefined expression like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not provide enough information to evaluate the limit directly. Indeterminate forms indicate that more analysis is needed, often requiring techniques like algebraic manipulation, factoring, or calculus-based approaches such as L'Hopital's Rule. L'Hopital's Rule is particularly useful for tackling indeterminate forms because it simplifies the expression by using derivatives, offering a way to potentially resolve or "determine" the limit.

Exploring Limits in Calculus

Limits describe the behavior of a function as the input approaches a certain point. They are foundational in calculus, used to define concepts like continuity, derivatives, and integrals. When a function's direct substitution yields an indeterminate form, we can't simply compute the limit using basic arithmetic. Instead, we interpret the limit as the value that the function approaches as we get extremely close to a specific input value. In graphical terms:

• The limit corresponds to where the y-values of the function tend to converge as the x-value approaches a particular point.
• If a function behaves erratically or has a gap, the limit reveals what value the function would attain in the ideal case.

Using limits, we see how functions behave at boundaries and unique points, enhancing our understanding of their overall behavior.

The Role of Derivatives in L'Hopital's Rule

Derivatives play a key role in L'Hopital's Rule, a powerful tool for evaluating limits. When faced with indeterminate forms like \(\frac{0}{0}\), L'Hopital's Rule instructs us to differentiate the numerator and denominator separately. This differentiation often simplifies the limit problem:

• The derivative of the numerator and the denominator transforms the original problem into a new expression, where the substitution should now yield a determinant form or clearer result.
• A successful use of L'Hopital’s Rule can reduce complex fractions to simpler expressions where direct substitution is feasible.

Ultimately, derivatives help by offering a straightforward alternate path to compute the desired limit, making complex calculus problems more accessible.