Business Calculus Exam 1 Review – Step-by-Step Study Guidance

Q1. Use a table of values to determine the following limits for $f(x) = (1 + \frac{1}{x})^x$:

• a. $\lim_{x \to -1^-} f(x)$

• b. $\lim_{x \to 0^+} f(x)$

• c. $\lim_{x \to \infty} f(x)$

Background

Topic: Limits and Continuity

This question tests your understanding of how to estimate limits using tables of values, especially for functions that are not easily simplified algebraically. It also introduces the concept of limits at infinity and one-sided limits.

Key Terms and Formulas

• Limit: The value that a function approaches as the input approaches a certain point.

• One-sided limit: The value the function approaches as $x$ approaches from one side (left or right).

• Limit at infinity: The value the function approaches as $x$ becomes very large (positive or negative).

Step-by-Step Guidance

1. For each part, choose values of $x$ that approach the target value from the correct side (e.g., for $x \to -1^-$, pick values just less than $-1$).

2. Calculate $f(x)$ for each chosen $x$ value and record the results in a table.

3. Observe the pattern in the function values as $x$ gets closer to the target value. Look for a trend (increasing, decreasing, approaching a specific number, etc.).

4. For the limit as $x \to \infty$, select large positive values of $x$ (e.g., $x = 10, 100, 1000$) and compute $f(x)$ for each.

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Q2. a. Identify any points of discontinuity for $f(x) = \frac{2x-10}{x^2-25}$, as well as the types of discontinuity.

Background

Topic: Limits and Continuity

This question asks you to find where a rational function is not continuous and to classify the type of discontinuity (removable, jump, or infinite/vertical asymptote).

Key Terms and Formulas

• Discontinuity: A point where the function is not continuous.

• Removable discontinuity: A 'hole' in the graph where the limit exists but the function is not defined or is defined differently.

• Vertical asymptote: Where the function grows without bound as $x$ approaches a certain value.

Step-by-Step Guidance

1. Factor the denominator: $x^2 - 25 = (x-5)(x+5)$.

2. Set the denominator equal to zero to find values where the function is undefined: $x-5=0$ and $x+5=0$.

3. Check if the numerator is also zero at these $x$-values to determine if the discontinuity is removable (hole) or non-removable (vertical asymptote).

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Q2. b. Find the limit as $x$ approaches $5$for$f(x)$ from part (a) above.

Background

Topic: Limits and Continuity

This question tests your ability to evaluate a limit at a point of discontinuity, possibly using algebraic simplification.

Key Terms and Formulas

• Limit at a point of discontinuity: Sometimes, even if a function is not defined at a point, the limit may exist.

Step-by-Step Guidance

1. Substitute $x=5$ into the denominator and numerator to check if you get $0/0$ (an indeterminate form).

2. If you get $0/0$, factor and simplify the expression to see if the limit can be found.

3. After simplifying, substitute $x=5$ into the simplified expression to find the limit.

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Q3. Find the following limits at infinity:

• a. $\lim_{x \to \infty} \frac{3x^4 - 2x^3}{15x^4 + 10}$

• b. $\lim_{x \to \infty} \frac{3x^3 - 2x^2}{15x^4 + 10}$

• c. $\lim_{x \to \infty} \frac{3x^4 - 2x^3}{15x^3 + 10}$

Background

Topic: Limits at Infinity

This question tests your understanding of how rational functions behave as $x$ becomes very large. The degrees of the numerator and denominator are important for determining the limit.

Key Terms and Formulas

• Degree of a polynomial: The highest power of $x$ in the expression.

• Limits at infinity for rational functions:

  • If degrees are equal: limit is the ratio of leading coefficients.

  • If numerator degree < denominator degree: limit is $0$.

  • If numerator degree > denominator degree: limit is $\infty$ or $-\infty$ (depending on sign).

Step-by-Step Guidance

1. Identify the degree of the numerator and denominator for each part.

2. Compare the degrees to determine which rule applies (see above).

3. If degrees are equal, divide the leading coefficients to find the limit.

4. If degrees are not equal, determine if the limit is $0$or$\infty$ based on which degree is higher.

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Q4. Find the value of $k$ that would make the piecewise-defined function continuous everywhere:

$\displaystyle f(x) = \begin{cases} \frac{2x^2 - 4x - 16}{x-4} & \text{if } x \neq 4 \\ kx-8 & \text{if } x = 4 \end{cases}$

Background

Topic: Continuity of Piecewise Functions

This question tests your ability to ensure a piecewise function is continuous at the point where the definition changes by matching the limit from both sides to the function value at that point.

Key Terms and Formulas

• Continuous at $x=a$: $\lim_{x \to a} f(x) = f(a)$

• Piecewise function: A function defined by different expressions for different intervals of $x$.

Step-by-Step Guidance

1. Find $\lim_{x \to 4} f(x)$ using the first piece ($x \neq 4$). If direct substitution gives $0/0$, factor and simplify.

2. Set the limit equal to the value of the function at $x=4$, which is $kx-8$ with $x=4$.

3. Solve for $k$ so that the function is continuous at $x=4$.

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