Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
1. Limits and Continuity
Introduction to Limits
Problem 2.5.50
Textbook Question
Determine limx→∞f(x) and limx→−∞f(x) for the following functions. Then give the horizontal asymptotes of f (if any).
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1
First, let's analyze the function f(x) = 4x(3x - \sqrt{9x^2 + 1}). We need to find the limits as x approaches infinity and negative infinity.
To find \lim_{x \to \infty} f(x), factor out x from the square root in the expression \sqrt{9x^2 + 1}. This gives us \sqrt{x^2(9 + \frac{1}{x^2})} = x\sqrt{9 + \frac{1}{x^2}}.
Rewrite the function as f(x) = 4x(3x - x\sqrt{9 + \frac{1}{x^2}}) = 4x^2(3 - \sqrt{9 + \frac{1}{x^2}}).
As x approaches infinity, \frac{1}{x^2} approaches 0, so \sqrt{9 + \frac{1}{x^2}} approaches \sqrt{9} = 3. Therefore, the expression 3 - \sqrt{9 + \frac{1}{x^2}} approaches 0.
Thus, \lim_{x \to \infty} f(x) = 4x^2 \cdot 0 = 0. Similarly, for \lim_{x \to -\infty} f(x), the same reasoning applies, leading to the conclusion that the horizontal asymptote is y = 0.
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