Lesson 7 (09/09/2026)

Lesson 7 (09/09/2026)

Today:

• Limits at infinity
• Horizontal Asymptote
• Slant Asymptotes

Office Hours: Monday, Wednesday, Friday 1:30pm – 2:30pm

Announcements:

• Exam 1 on Wednesday 09/23, 8pm – 9pm
  • → Study Guide, instructions, Seating Chart will be posted tomorrow.
• HW7 Due: Thursday 09/10
• Quiz 4 (Lesson 6): Thursday 09/10
• Feasting with Faculty (TODAY!! 6pm – 7pm Windsor)

Review example: Determine the vertical asymptotes of

\[ f(x) = \frac{(x-1)^2 (x+3)}{(x-1)(x+3)^2 (x-7)} \]

Possible candidates: \( x = 1 \), \( x = -3 \), \( x = 7 \)

\( \underline{x=1} \):

\[ \lim_{x \to 1} \frac{(x-1)^2 (x+3)}{(x-1)(x+3)^2 (x-7)} = \lim_{x \to 1} \frac{(x-1)(x+3)}{(x+3)^2 (x-7)} = \frac{(0)(3)}{3^2 (-7)} = 0 \]

has a hole at \( x = 1 \)

but when \( x \to -3 \), \( x \to 7 \) still have a term that goes to \( 0 \) in denominator

\( x = -3 \), \( x = 7 \) are two V.A

Evaluate \( \lim_{x \to a} \frac{1}{x} \)

\( a = 0 \leadsto \) Last Lecture: \[ \lim_{x \to 0^+} \frac{1}{x} = +\infty, \quad \lim_{x \to 0^-} \frac{1}{x} = -\infty \]

\( a \neq 0 \leadsto \) we can plug in: \[ \lim_{x \to a} \frac{1}{x} = \frac{1}{a} \]

\[ \lim_{x \to 10} \frac{1}{x} = \frac{1}{10} = 0.1 \qquad\qquad \lim_{x \to -10} \frac{1}{x} = -0.1 \]

\[ \lim_{x \to 1000} \frac{1}{x} = \frac{1}{1000} = 0.001 \qquad\qquad \lim_{x \to -100} \frac{1}{x} = -0.01 \]

\[ \lim_{x \to 1000000} \frac{1}{x} = 0.000001 \qquad\qquad \lim_{x \to -1000000} \frac{1}{x} = -0.000001 \]

As \( x \) gets LARGE AND positive, \( \frac{1}{x} \) gets closer to \( 0 \).

Notation:

\[ \lim_{x \to +\infty} \frac{1}{x} = 0 \]

\[ \lim_{x \to -\infty} \frac{1}{x} = 0 \]

Graph of \(f(x) = \frac{1}{x}\)

\(y=0\) is a Horizontal Asymptote

\(x=0\) is a Vertical Asymptote

Horizontal Asymptote:

\( y = L \) (a finite number) is a Horizontal asymptote to \( f(x) \)

if

\[ \lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L \]

Any function \( f(x) \) can have atmost 2 Horizontal Asymptotes

as \(x\) gets LARGE \(\frac{1}{x^2}\) also goes to 0

in general

\[ \lim_{x \to \infty} \frac{c}{x^n} = 0 \] \[ \lim_{x \to -\infty} \frac{c}{x^n} = 0 \]

for any \(n > 0\) and any constant \(C\).

Example:

\[ \lim_{x \to \infty} \frac{5x^2+3}{8x^2-7x+2} \]

Divide both numerator, denominator by \(x^{\text{highest degree}}\).

here, divide with \(x^2\)

\[ \lim_{x \to \infty} \frac{5x^2+3}{8x^2-7x+2} = \lim_{x \to \infty} \frac{\frac{5x^2+3}{x^2}}{\frac{8x^2-7x+2}{x^2}} = \lim_{x \to \infty} \frac{5 + \frac{3}{x^2}^{\to 0}}{8 - \frac{7}{x}_{\to 0} + \frac{2}{x^2}^{\to 0}} = \frac{5}{8} \]

\[ \lim_{x \to -\infty} \frac{5x^2+3}{8x^2-7x+2} = \lim_{x \to -\infty} \frac{\frac{5x^2+3}{x^2}}{\frac{8x^2-7x+2}{x^2}} = \lim_{x \to -\infty} \frac{5 + \frac{3}{x^2}^{\to 0}}{8 - \frac{7}{x}_{\to 0} + \frac{2}{x^2}^{\to 0}} = \frac{5}{8} \]

\(y = 5/8\) is the only Horizontal Asymptote

eg 1

\[ \lim_{x \to \infty} \frac{x+8}{x^2-11} \]

Divide with \(x^2\) in Numerator & Denominator

\[ \lim_{x \to \infty} \frac{\frac{x+8}{x^2}}{\frac{x^2-11}{x^2}} = \lim_{x \to \infty} \frac{\frac{1}{x} + \frac{8}{x^2}}{1 - \frac{11}{x^2}} = 0 \]

Similarly,

\[ \lim_{x \to -\infty} \frac{x+8}{x^2-11} = 0 \]

\(y = 0\) is the only Horizontal Asymptote

Example 7

\[ \lim_{x \to +\infty} \frac{x^3 + 8}{x^2 - 11} \]

Divide with \( x^2 \)

\[ \lim_{x \to +\infty} \frac{\frac{x^3 + 8}{x^2}}{\frac{x^2 - 11}{x^2}} = \lim_{x \to +\infty} \frac{x + \frac{8}{x^2} \to 0}{1 - \frac{11}{x^2} \to 0} \]

\[ = \lim_{x \to +\infty} x = +\infty \]

\[ \lim_{x \to -\infty} \frac{x^3 + 8}{x^2 - 11} \quad \text{divide with } x^3 \]

\[ \lim_{x \to -\infty} \frac{\frac{x^3 + 8}{x^3}}{\frac{x^2 - 11}{x^3}} = \lim_{x \to -\infty} \frac{1 + \frac{8}{x^3} \to 0}{\frac{1}{x} \to 0 - \frac{11}{x^3} \to 0} = -\infty \]

No Horizontal Asymptote

Slant Asymptotes

If \( \lim_{x \to \pm\infty} f(x) = \pm\infty \), it could have a slant asymptote.

To find them, do polynomial long division:

\[ \frac{x^3+8}{x^2-11} \rightsquigarrow \]

\[ \begin{array}{r@{}c@{\;}r@{\;}r@{\;}r@{\;}r} & & x & & & \\ \hline x^2+0x-11 & \big) & x^3 & +0x^2 & +0x & +8 \\ & & x^3 & +0x^2 & -11x & \\ \hline & & & & 11x & +8 \end{array} \]

\[ \frac{x^3+8}{x^2-11} = x + \frac{11x+8}{x^2-11} \quad \text{goes to } 0 \text{ as } x \to \infty \text{ or } x \to -\infty \]

As \( x \to \pm\infty \):

\[ \frac{x^3+8}{x^2-11} \approx x \rightsquigarrow y = x \text{ is a slant asymptote.} \]

Example:

\[ \lim_{x \to \infty} \frac{4x^2 - x - 5}{x - 1} \]

Polynomial long division:

\[ \begin{array}{r@{}c@{\;}r@{\;}r@{\;}r} & & 4x & +3 & \\ \hline x-1 & \big) & 4x^2 & -x & -5 \\ & & 4x^2 & -4x & \\ \hline & & & 3x & -5 \\ & & & 3x & -3 \\ \hline & & & & -2 \end{array} \]

\[ \frac{4x^2 - x - 5}{x - 1} = 4x + 3 - \frac{2}{x - 1} \]

As \( x \to \pm\infty \), \( \frac{2}{x-1} \to 0 \).

\( y = 4x + 3 \) is a Slant Asymptote

Eg:

\[ \lim_{x \to \infty} \frac{5x + 2}{3x + \sqrt{4x^2 + 7}} \]

divide with \(x\)

\[ \lim_{x \to \infty} \frac{\frac{5x+2}{x}}{\frac{3x + \sqrt{4x^2+7}}{x}} = \lim_{x \to \infty} \frac{5 + \frac{2}{x}}{3 + \frac{\sqrt{4x^2+7}}{x}} \]

\(x > 0 \implies x = \sqrt{x^2}\)

\[ = \lim_{x \to \infty} \frac{5 + \frac{2}{x}^{\to 0}}{3 + \sqrt{4 + \frac{7}{x^2}^{\to 0}}} \] \[ = \frac{5}{3 + \sqrt{4}} = \frac{5}{3 + 2} = 1. \]

\(y = 1\) is a H.A.

If \( x < 0 \implies x = -\sqrt{x^2} \)

Be careful when you take \( \lim_{x \to -\infty} \)

\[ \lim_{x \to -\infty} \frac{5x + 2}{3x + \sqrt{4x^2 + 7}} = \lim_{x \to -\infty} \frac{5 + \frac{2}{x}}{3 + \frac{\sqrt{4x^2 + 7}}{x}} \]

\( x = -\sqrt{x^2} \)

\[ = \lim_{x \to -\infty} \frac{5 + \frac{2}{x}}{3 - \sqrt{4 + \frac{7}{x^2}}} \] \[ = \frac{5}{3 - 2} = 5 \]

\( y = 5 \) is a H.A.