Problem 55 - Thomas Calculus

Question

Find the limits in Exercises \(53-58.\) $$\lim \left(\frac{x^{2}}{2}-\frac{1}{x}\right)$$ \begin{equation} \begin{array}{ll}{\text { a. }} & {x \rightarrow 0^{+}} & {\text { b. } x \rightarrow 0^{-}} \\ {\text { c. }} & {x \rightarrow \sqrt[3]{2}} & {\text { d. } x \rightarrow-1}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. \( -\infty \); b. \( \infty \); c. finite; d. \( \frac{3}{2} \).

Step 1: Understanding the Expression

The expression given is \( \frac{x^2}{2} - \frac{1}{x} \). Our goal is to evaluate this limit under various conditions for \( x \): as \( x \to 0^+ \), \( x \to 0^- \), \( x \to \sqrt[3]{2} \), and \( x \to -1 \).

Step 2: Find the Limit as \( x\to0^+ \)

As \( x \to 0^+ \), the term \( \frac{1}{x} \to +\infty \). Hence, the expression \( \frac{x^2}{2} - \frac{1}{x} \to -\infty \), since \( \frac{1}{x} \) dominates and is negative.

Step 3: Find the Limit as \( x\to0^- \)

As \( x \to 0^- \), the term \( \frac{1}{x} \to -\infty \). Again, \( \frac{x^2}{2} - \frac{1}{x} \to +\infty \) because \( \frac{1}{x} \) dominates and is positive.

Step 4: Find the Limit as \( x\to\sqrt[3]{2} \)

Substitute \( x = \sqrt[3]{2} \) directly into the expression since the function is continuous at this point: \[ \frac{(\sqrt[3]{2})^2}{2} - \frac{1}{\sqrt[3]{2}} = \frac{2^{2/3}}{2} - 2^{-1/3}. \] Simplify this using properties of exponents.

Step 5: Simplified Expression for \( x\to\sqrt[3]{2} \)

Calculate \( 2^{2/3} \) and \( 2^{-1/3} \). The expression simplifies to: \[ \frac{2^{2/3}}{2} - 2^{-1/3} = \frac{\sqrt[3]{4}}{2} - \frac{1}{\sqrt[3]{2}}. \] Thus, the limit is a finite number.

Step 6: Find the Limit as \( x\to-1 \)

Substitute \( x = -1 \) directly into the expression since it is continuous at \( x = -1 \): \[ \frac{(-1)^2}{2} - \frac{1}{-1} = \frac{1}{2} + 1 = \frac{3}{2}. \]

Step 7: Conclusion: Consolidate Limits

The limits computed for each condition are: \( \lim_{x \to 0^+} -\infty \), \( \lim_{x \to 0^-} +\infty \), \( \lim_{x \to \sqrt[3]{2}} = \frac{2^{2/3}}{2} - 2^{-1/3} \), and \( \lim_{x \to -1} = \frac{3}{2} \).

Key Concepts

Continuous Functions

Continuous functions are fundamental in calculus, representing functions without gaps, jumps, or abrupt changes within a specific interval. A function is continuous at a point if you can draw it without lifting your pencil from the paper in its graphical form. Mathematically, a function \( f(x) \) is continuous at a point \( x = a \) if the limit as \( x \) approaches \( a \) equals the function value at \( a \), i.e., \( \lim_{x \to a} f(x) = f(a) \).

Recognizing a continuous function allows students to substitute variables directly into the function when finding limits. In the given exercise, we dealt with continuous functions at points like \( x = \sqrt[3]{2} \) and \( x = -1 \). By direct substitution, we find the limit without the need for approximation or additional steps. This property simplifies our work significantly.
Continuous functions assure that minor changes in the input do not cause drastic swings in the function's output, supporting deeper analyses such as finding limits, derivatives, and integrals.

Limits at Infinity

In calculus, limits at infinity deal with the behavior of a function as the independent variable approaches infinity or negative infinity. This concept helps in understanding how functions behave when the values of \( x \) grow very large or very small, giving insights into the end behavior of the function.

For the expression in the exercise, particularly when analyzing \( x \to 0^+ \) and \( x \to 0^- \), we encounter infinite limits. The term \( \frac{1}{x} \) plays a crucial role as it tends to \( +\infty \) or \( -\infty \), depending on whether \( x \) approaches zero from the positive or negative side, respectively. This tells us about the vertical asymptotic behavior of the function near zero.

• As \( x \to 0^+ \): \( \frac{1}{x} \to +\infty \), so the expression diverges to \( -\infty \).
• As \( x \to 0^- \): \( \frac{1}{x} \to -\infty \), causing the expression to diverge to \( +\infty \).

Limits at infinity are invaluable for visualizing the growth, decay, and asymptotic tendencies of functions.

Properties of Exponents

The properties of exponents are a cornerstone for solving complex algebraic and calculus problems. These properties help simplify expressions and are crucial for evaluating limits and other mathematical operations. The basic exponent rules include:

• \( a^{m} \times a^{n} = a^{m+n} \)
• \( (a^{m})^{n} = a^{m \times n} \)
• \( a^{0} = 1 \) for any \( a eq 0 \)
• \( a^{-n} = \frac{1}{a^{n}} \)

In the given exercise, exponent properties are particularly useful when substituting \( x = \sqrt[3]{2} \). The expression involves simplifying terms like \( 2^{2/3} \) and \( 2^{-1/3} \), which use the negative exponent rule to express roots as fractions. This simplification provides a clear path to solving for the limit by transforming complex fractional exponents into more manageable numbers.

Understanding these exponent rules allows us to transform and evaluate expressions efficiently, revealing the true behavior of the function under the limit.