Limits at Infinity and Review of Limit Techniques

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Limits at Infinity: End Behavior

• Definition: These limits address what happens to a function as the independent variable $(x)$ approaches positive infinity $(x \to \infty)$ or negative infinity $(x \to -\infty)$. They are referred to as "end behavior limits" because they describe the function's behavior at the extreme ends of the number system.

• Connection to Pre-Calculus: This concept is directly related to finding horizontal asymptotes, as studied in pre-calculus.

• Methodology: Instead of specific points, we analyze the function's behavior by imagining plugging in extremely large positive or extremely large negative numbers into the equation.

• Possible Outcomes: There are generally two main results for limits at infinity:

  1. Approaching a Constant: The function "calms down" to a specific finite value (e.g., $0$, $-6$, $200$). This value represents a horizontal asymptote.

  2. Approaching Infinity: The function "takes off" indefinitely towards positive infinity $(+\infty)$ or negative infinity $(-\infty)$, meaning it never settles to a finite value.

• Goal: Determine whether the function calms down to a constant or grows/shrinks without bound toward infinity.

• End Behavior Equation: Sometimes, we're interested in the equation that the function's ends are following. For example, a complex rational function might ultimately follow a simple constant function (e.g., $y=3$) or a polynomial (e.g., $y = \frac{3}{4}x^3$) at its extremes.

Analyzing Limits of Rational Polynomials at Infinity

• General Principle: Limits at infinity for rational polynomials $\left(\frac{P(x)}{Q(x)}\right)$ can often be determined by comparing the degrees of the numerator and denominator.

• Case 1: Same Degree ($P(x)$ and $Q(x)$ have the same highest power, $n$)

  • Result: The limit will always be a constant, specifically the ratio of the leading coefficients of the highest degree terms.

  • Formula: If $\lim_{x \to \pm\infty} \frac{ax^n + \dots}{bx^n + \dots}$, the limit is $\frac{a}{b}$.

  • Method: Divide every term in the numerator and denominator by the highest power of $x$ found in the denominator (i.e., $x^n$). As $x \to \pm\infty$, any term with $x$ in the denominator (like $\frac{\text{constant}}{x^k}$ where k > 0) will approach $0$. The remaining constant terms will give the limit.

  • Example: $\lim<em>{x \to \pm\infty} \frac{3x^2 - x + 5}{4x^2 + 2x - 1}$. Degrees are both $2$. Divide by $x^2$: $\lim</em>{x \to \pm\infty} \frac{3 - \frac{1}{x} + \frac{5}{x^2}}{4 + \frac{2}{x} - \frac{1}{x^2}} = \frac{3 - 0 + 0}{4 + 0 - 0} = \frac{3}{4}$.

• Case 2: Higher Degree in the Denominator ($Q(x)$ has a higher power than $P(x)$, n > m)

  • Result: The limit will always be $0$. This corresponds to a horizontal asymptote at $y=0$.

  • Formula: If $\lim_{x \to \pm\infty} \frac{ax^m + \dots}{bx^n + \dots}$ where m < n, the limit is $0$.

  • Method: Divide every term by the highest power of $x$ in the denominator (i.e., $x^n$). All terms will have $x$ in their denominator, causing them to approach $0$. The numerator will reduce to $0$ and the denominator to a constant.

  • Example: $\lim<em>{x \to \pm\infty} \frac{5x^3 - 2x^2}{6x^4 - 8x}$. Denominator degree ($4$) is higher than numerator degree ($3$). Divide by $x^4$: $\lim</em>{x \to \pm\infty} \frac{\frac{5}{x} - \frac{2}{x^2}}{6 - \frac{8}{x^3}} = \frac{0 - 0}{6 - 0} = 0$.

• Case 3: Higher Degree in the Numerator ($P(x)$ has a higher power than $Q(x)$, m > n)

  • Result: The limit will always be positive infinity $(+\infty)$ or negative infinity $(-\infty)$. There is no horizontal asymptote.

  • Formula: If $\lim_{x \to \pm\infty} \frac{ax^m + \dots}{bx^n + \dots}$ where m > n, the limit is $\pm\infty$.

  • Method: Divide every term by the highest power of $x$ in the denominator (i.e., $x^n$). This will leave some $x$ terms in the numerator. The behavior of these remaining $x$ terms (especially the highest power) will determine if the function goes to $\pm\infty$.

  • End Behavior Equation: The function's end behavior will follow a polynomial curve. This equation can be found by polynomial division or by simply taking the ratio of the highest degree terms, simplifying, and including any remaining constant offset.

  • Example: $\lim_{x \to \pm\infty} \frac{3x^5 - 2x^2 + 1}{4x^2 + 5x + 2}$. Numerator degree ($5$) is higher than denominator degree ($2$). The end behavior equation is approximately $\frac{3x^5}{4x^2} = \frac{3}{4}x^3$. For large positive $x$, it's $+\infty$. For large negative $x$, it's $-\infty$. The precise end behavior equation often keeps more terms after division.

  • Example: $\lim<em>{x \to \pm\infty} \frac{x^3 + 2x^2 + 1}{x^2 + 1}$. Divide by $x^2$: $\lim</em>{x \to \pm\infty} \frac{x + 2 + \frac{1}{x^2}}{1 + \frac{1}{x^2}} = x + 2$. The end behavior equation is $y = x+2$. For large positive $x$, it's $+\infty$. For large negative $x$, it's $-\infty$.

Limits Involving Trigonometric Functions at Infinity

• Principle: When a trigonometric function (like sine or cosine) whose output is bounded (between $-1$ and $1$) is divided by a term that approaches infinity, the overall expression will approach $0$. This is often proven using the Squeeze Theorem.

• Example: $\lim_{x \to \pm\infty} \frac{2x + \sin x}{5x + \cos x}$

  • Divide all terms by $\frac{1}{x}$ (highest power in denominator that includes $x$).

  • This yields $\lim_{x \to \pm\infty} \frac{2 + \frac{\sin x}{x}}{5 + \frac{\cos x}{x}}$.

  • Analysis of $\lim<em>{x \to \pm\infty} \frac{\sin x}{x}$ and $\lim</em>{x \to \pm\infty} \frac{\cos x}{x}$:

    • The functions $\sin x$ and $\cos x$ always output values between $-1$ and $1$ ($-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$).

    • As $x$ approaches $\pm\infty$, the denominator $x$ becomes infinitely large while the numerator remains bounded. Therefore, both fractions approach $0$.

  • Squeeze Theorem Proof for $\lim_{x \to \pm\infty} \frac{\sin x}{x}$:

    • We know $-1 \le \sin x \le 1$.

    • Divide by $x$ (assuming x > 0; the logic extends to x < 0): $\frac{-1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$.

    • As $x \to \infty$, $\lim<em>{x \to \infty} \frac{-1}{x} = 0$ and $\lim</em>{x \to \infty} \frac{1}{x} = 0$.

    • By the Squeeze Theorem, $\lim_{x \to \infty} \frac{\sin x}{x} = 0$. The same applies for $\cos x$ and $\text{as } x \to -\infty$.

  • Result: Substituting these limits, we get $\frac{2 + 0}{5 + 0} = \frac{2}{5}$.

Connection to Pre-Calculus Concepts (Review of Limits for Graph Analysis)

• Vertical Asymptotes (VA): Represent locations where the limit does not exist (DNE), often because the function approaches $\pm\infty$ from a specific side. Mathematically, it's a point where the denominator is zero, and the factor does not cancel with the numerator.

• Roots (x-intercepts): Points where $f(x)=0$. These are places where direct substitution for the limit works perfectly, yielding zero.

• Holes: Occur when a factor in the denominator cancels with a factor in the numerator, leading to an indeterminate form of $0/0$ for the limit at that point. The limit exists and is a finite number, representing the $y$-coordinate of the hole, but the function is undefined at that specific $x$-value.

• Horizontal Asymptotes (HA): Equivalent to end behavior limits (limits as $x \to \pm\infty$) that result in a constant value.

• Example: Analyzing $\frac{2x^2-4x-6}{3x^2-27}$

  • Simplification: Factor the numerator and denominator.

    • Numerator: $2(x^2 - 2x - 3) = 2(x - 3)(x + 1)$

    • Denominator: $3(x^2 - 9) = 3(x - 3)(x + 3)$

  • Simplified function: $\frac{2(x + 1)}{3(x + 3)}$ (after canceling $(x-3)$).

  • Horizontal Asymptote / End Behavior: Since the highest degree in both numerator ($x^2$) and denominator ($x^2$) is the same, the limit as $x \to \pm\infty$ is the ratio of leading coefficients: $\frac{2}{3}$. This is the HA.

  • X-intercepts: From the numerator of the simplified function, set $2(x + 1) = 0 \implies x = -1$.

  • Vertical Asymptote: From the denominator of the simplified function, set $3(x + 3) = 0 \implies x = -3$.

  • Hole: The factor $(x-3)$ canceled out, indicating a hole at $x=3$. To find the $y$-coordinate of the hole, evaluate the limit of the simplified function as $x \to 3$: $\lim_{x \to 3} \frac{2(x + 1)}{3(x + 3)} = \frac{2(3 + 1)}{3(3 + 3)} = \frac{2(4)}{3(6)} = \frac{8}{18} = \frac{4}{9}$. So, there's a hole at $(3, \frac{4}{9})$.

Special and Tricky Limits

• Limit of $\sin(\frac{1}{x})$ as $x \to \pm\infty$:

  • As $x \to \pm\infty$, the argument $\frac{1}{x}$ approaches $0$.

  • Therefore, $\lim_{x \to \pm\infty} \sin(\frac{1}{x}) = \sin(0) = 0$.

• Limit of $x \sin(\frac{1}{x})$ as $x \to \pm\infty$ (Indeterminate Form $\infty \cdot 0$):

  • This is an indeterminate form that requires a substitution technique.

  • Substitution: Let $y = \frac{1}{x}$.

    • As $x \to \pm\infty$, $\frac{1}{x} \to 0$, so $y \to 0$.

    • Also, if $y = \frac{1}{x}$, then $x = \frac{1}{y}$.

  • Rewrite the limit: Substitute $y$ into the expression:

    • $\lim<em>{x \to \pm\infty} x \sin(\frac{1}{x}) = \lim</em>{y \to 0} \frac{1}{y} \sin(y) = \lim_{y \to 0} \frac{\sin y}{y}$.

  • Apply known trigonometric limit: The limit as $y \to 0$ of $\frac{\sin y}{y}$ is a fundamental trigonometric limit equal to $1$.

  • Result: $\lim_{x \to \pm\infty} x \sin(\frac{1}{x}) = 1$.

Review of Trigonometric Substitution for Limits at a Point

• Problem: Evaluate $\lim_{\theta \to \frac{\pi}{4}} \frac{\cos(\frac{3\pi}{4} - \theta)}{\sin(\theta - \frac{\pi}{4})}$ (Direct substitution yields the indeterminate form $0/0$).

• Technique: Substitution to make the limit approach zero.

  • Let $x = \theta - \frac{\pi}{4}$. As $\theta \to \frac{\pi}{4}$, $x \to 0$.

  • From $x = \theta - \frac{\pi}{4}$, we can express $\theta = x + \frac{\pi}{4}$.

• Rewrite the expression in terms of $x$:

  • Denominator: $\sin(\theta - \frac{\pi}{4}) = \sin(x)$.

  • Numerator: $\cos(\frac{3\pi}{4} - \theta) = \cos(\frac{3\pi}{4} - (x + \frac{\pi}{4})) = \cos(\frac{3\pi}{4} - x - \frac{\pi}{4}) = \cos(\frac{2\pi}{4} - x) = \cos(\frac{\pi}{2} - x)$.

• Apply Trigonometric Identities:

  • Recall the cosine subtraction identity: $\cos(A - B) = \cos A \cos B + \sin A \sin B$.

  • So, $\cos(\frac{\pi}{2} - x) = \cos(\frac{\pi}{2})\cos(x) + \sin(\frac{\pi}{2})\sin(x)$.

  • Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$, this simplifies to $(0)\cos(x) + (1)\sin(x) = \sin(x)$.

  • (Alternatively, use the cofunction identity: $\cos(\frac{\pi}{2} - x) = \sin(x)$).

• Evaluate the new limit:

  • The limit becomes $\lim_{x \to 0} \frac{\sin x}{\sin x}$.

  • For $x \ne 0$, $\frac{\sin x}{\sin x} = 1$.

  • Result: $\lim_{x \to 0} 1 = 1$.