Welcome back to Professor Baker's Math Class! In this section, we are shifting our focus from looking at specific points on a graph (like we did with vertical asymptotes) to looking at the "end behavior" of functions. Specifically, we are exploring Section 2-6: Limits at Infinity and Horizontal Asymptotes.

What happens to the $y$-value of a function as $x$ gets infinitely large ($x \to \infty$) or infinitely negative ($x \to -\infty$)? Let's dive into the rules and techniques needed to solve these problems.

1. The Definition of a Horizontal Asymptote

A horizontal asymptote describes the behavior of a graph at the far ends of the $x$-axis. We define the line $y = L$ as a horizontal asymptote of the curve $y = f(x)$ if either:

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

Essentially, if the function flattens out and approaches a specific number as you go further to the right or left, that number is your horizontal asymptote.

2. Rational Functions: The Shortcuts

When dealing with rational functions (a polynomial divided by a polynomial), compare the degree (highest exponent) of the numerator to the degree of the denominator. Based on the notes, here is the quick guide:

  • Bottom Heavy (Degree of Denom > Numerator): The limit is $0$. Therefore, the horizontal asymptote is $y = 0$.
    Example: $f(x) = \frac{3x}{x^2+5} \to 0$.
  • Equal Degrees: The limit is the ratio of the leading coefficients.
    Example: $h(x) = \frac{5x^3 - 2x + 7}{6x^3 + 5x^2 - 2} \to \frac{5}{6}$. The asymptote is $y = \frac{5}{6}$.
  • Top Heavy (Degree of Numerator > Denom): The limit is $\infty$ or $-\infty$. There is no horizontal asymptote.

3. The Algebraic Method

While the shortcuts are great, you often need to show your work or handle non-polynomial functions. The standard technique is to divide every term by the highest power of $x$ in the denominator.

This relies on the fundamental theorem:

$$\lim_{x \to \infty} \frac{1}{x^r} = 0 \quad \text{where } r > 0$$

For example, evaluating $\lim_{x \to \infty} \frac{3x^2 - x - 2}{5x^2 + 4x + 1}$, we divide everything by $x^2$. This turns all the smaller terms (like $\frac{x}{x^2}$ or $\frac{2}{x^2}$) into zero, leaving us with just the coefficients $\frac{3}{5}$.

4. Dealing with Radicals and Conjugates

Things get trickier when square roots are involved. There are two main scenarios to watch out for:

Scenario A: The Conjugate Method
If you have a limit limit like $\lim_{x \to \infty} (\sqrt{x^2+1} - x)$, simply plugging in infinity gives you $\infty - \infty$, which is an indeterminate form. To solve this, multiply the numerator and denominator by the conjugate $(\sqrt{x^2+1} + x)$. This eliminates the root in the numerator and allows you to solve the limit.

Scenario B: Negative Infinity Trap
Be very careful when $x \to -\infty$ with radicals! Remember that $\sqrt{x^2} = |x|$.

  • If $x > 0$, then $\sqrt{x^2} = x$.
  • If $x < 0$, then $\sqrt{x^2} = -x$.
When evaluating a limit like $\lim_{x \to -\infty} \frac{\sqrt{4x^2+3x}}{x}$, pulling the $x^2$ out of the square root introduces a negative sign because we are approaching negative infinity. The result would be $-\sqrt{4} = -2$, not positive 2.

5. Transcendental Functions and The Squeeze Theorem

Don't forget your trig and inverse trig functions!

  • $\lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2}$
  • $\lim_{x \to \infty} \sin(x) = \text{DNE}$ (because it oscillates forever)

However, if an oscillating function is multiplied by a decaying function, like $\lim_{x \to \infty} e^{-2x}\cos(x)$, we use the Squeeze Theorem. Since $\cos(x)$ is bounded between -1 and 1, and $e^{-2x}$ goes to 0, the whole function is "squeezed" to 0.

Keep practicing these algebraic manipulations, and always check the sign of infinity!