Limits at Infinity and Review of Limit Techniques
Announcements
Homecoming Dance: Tickets for the dance on October 4 are available in the main office until October 3 at midnight. Student council will sell tickets during lunches and answer questions.
Homecoming Spirit Week: Dress-up days will be posted around the school next week.
Powder Puff Team: Girls interested in joining their grade's powder puff team can sign up in the athletic office until September 23. Games are in October with a 10 entrance fee.
Travel Night: An informational session for upcoming educational experiences in Northern Italy, Ecuador/Galapagos Islands, Japan, or Iceland will be held on Wednesday, September 24, at 6 PM in Bayport's Forum Room. Sign-up details are on Schoology.
Choice Pirate Time: On Wednesday and Thursday this week, students must log into Flex to make a selection. Options include Bayport FFA's new member sign-up meeting on Wednesday, September 24, in Room A114.
YSL Club: Join the Schoology group using the code posted in Schoology. The first all-member meeting is Thursday, September 25, during Pirate Time in the PAC. Sign up in Flex and join the Schoology group for updates.
Mindfulness Monday Task: For the next time you walk, send positive mental well wishes to everyone who crosses your path, focusing on shared humanity and wishing them good health and happiness.
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:
Approaching a Constant: The function "calms down" to a specific finite value (e.g., 0, -6, 200). This value represents a horizontal asymptote.
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{x \to \pm\infty} \frac{3x^2 - x + 5}{4x^2 + 2x - 1}. Degrees are both 2. Divide by x^2: \lim{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{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{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{x \to \pm\infty} \frac{x^3 + 2x^2 + 1}{x^2 + 1}. Divide by x^2: \lim{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{x \to \pm\infty} \frac{\sin x}{x} and \lim{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{x \to \infty} \frac{-1}{x} = 0 and \lim{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{x \to \pm\infty} x \sin(\frac{1}{x}) = \lim{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.