Find the vertical asymptotes of \(f(x) = \frac{x}{x^2(x-7)(x+3)(x-7)}\).
Solution:
Vertical asymptotes occur when the denominator equals zero but the numerator doesn't.
Setting denominator = 0: \(x = 1, x = -3, x = 7\)
Check \(x = 1\): Since \(\lim_{x \to 1} f(x) = 0\), this is NOT a vertical asymptote.
Check \(x = -3\) and \(x = 7\): These make the denominator very small while the numerator approaches a finite value, creating vertical asymptotes.
Answer: \(x = -3\) and \(x = 7\) are vertical asymptotes.
Notation:
Basic Limit Rules: Let \(c\) be a constant and \(n > 0\):
\[\lim_{x \to \infty} \frac{c}{x^n} = 0 \quad \text{and} \quad \lim_{x \to -\infty} \frac{c}{x^n} = 0\]For \(c = 7, n = 2\): \(\frac{7}{x^2}\)
Definition: The line \(y = L\) is a horizontal asymptote of \(y = f(x)\) if:
\[\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L\]In general, any function \(f(x)\) has 0, 1, or 2 horizontal asymptotes.
Find \(\lim_{x \to \infty} \frac{5x^2 + 3}{8x^2 - 7x + 2}\) and \(\lim_{x \to -\infty} \frac{5x^2 + 3}{8x^2 - 7x + 2}\)
Solution:
Degree of numerator = Degree of denominator = 2
Divide both numerator and denominator by \(x^2\):
\[\lim_{x \to \infty} \frac{5x^2 + 3}{8x^2 - 7x + 2} = \lim_{x \to \infty} \frac{5 + \frac{3}{x^2}}{8 - \frac{7}{x} + \frac{2}{x^2}} = \frac{5 + 0}{8 - 0 + 0} = \frac{5}{8}\]Similarly, \(\lim_{x \to -\infty} \frac{5x^2 + 3}{8x^2 - 7x + 2} = \frac{5}{8}\)
Horizontal asymptote: \(y = \frac{5}{8}\)
Find \(\lim_{x \to \infty} \frac{x + 1}{8x^2}\) and \(\lim_{x \to -\infty} \frac{x + 1}{8x^2}\)
Solution:
Degree of numerator = 1, Degree of denominator = 2
Divide by \(x^2\):
\[\lim_{x \to \infty} \frac{x + 1}{8x^2} = \lim_{x \to \infty} \frac{\frac{1}{x} + \frac{1}{x^2}}{8} = \frac{0 + 0}{8} = 0\]Similarly, \(\lim_{x \to -\infty} \frac{x + 1}{8x^2} = 0\)
Horizontal asymptote: \(y = 0\)
Find \(\lim_{x \to \infty} \frac{3x^2 + x}{x^2}\) and \(\lim_{x \to -\infty} \frac{3x^2 + x}{x^2}\)
Solution:
Degree of numerator = Degree of denominator = 2
Divide by \(x^2\):
\[\lim_{x \to \infty} \frac{3x^2 + x}{x^2} = \lim_{x \to \infty} \frac{3 + \frac{1}{x}}{1} = \frac{3 + 0}{1} = 3\]Similarly, \(\lim_{x \to -\infty} \frac{3x^2 + x}{x^2} = 3\)
Horizontal asymptote: \(y = 3\)
When do slant asymptotes occur?
When \(\lim_{x \to \infty} f(x) = \infty\) or \(-\infty\), the function \(f(x)\) can have a slant asymptote.
Find the slant asymptote of \(\frac{7x^3 - x}{3x^2 + 1}\)
Solution:
Use long division or divide by the highest power in denominator (\(x^2\)):
\[\frac{7x^3 - x}{3x^2 + 1} = \frac{7x - \frac{1}{x}}{3 + \frac{1}{x^2}}\]As \(x \to \infty\): \(\lim_{x \to \infty} \frac{7x - \frac{1}{x}}{3 + \frac{1}{x^2}} = \lim_{x \to \infty} \frac{7x}{3} = \infty\)
The slant asymptote is \(y = \frac{7x}{3}\)
Key Strategy: When finding limits at infinity for rational functions, always divide numerator and denominator by the highest power of \(x\) in the denominator.
Find \(\lim_{x \to \infty} \frac{\sqrt{x^2 + x}}{7x + 2}\)
Solution:
For \(x > 0\), \(\sqrt{x^2} = |x| = x\)
Factor out \(x\) from under the square root:
\[\sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} = x\sqrt{1 + \frac{1}{x}}\]Therefore:
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + x}}{7x + 2} = \lim_{x \to \infty} \frac{x\sqrt{1 + \frac{1}{x}}}{7x + 2} = \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x}}}{7 + \frac{2}{x}} = \frac{1}{7}\]Important Note: When \(x < 0\), \(\sqrt{x^2} = |x| = -x\), so the analysis changes for \(\lim_{x \to -\infty}\).
Find \(\lim_{x \to -\infty} \frac{\sqrt{x^2 + x}}{7x + 2}\)
Solution:
For \(x < 0\), \(\sqrt{x^2} = |x| = -x\)
Factor out \(x\) from under the square root:
\[\sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} = -x\sqrt{1 + \frac{1}{x}}\]Therefore:
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + x}}{7x + 2} = \lim_{x \to \infty} \frac{-x\sqrt{1 + \frac{1}{x}}}{7x + 2} = \lim_{x \to \infty} \frac{-\sqrt{1 + \frac{1}{x}}}{7 + \frac{2}{x}} = -\frac{1}{7}\]