More with continuity; limits with infinity (Calc 1; Lecture 1-5; Fall 26)

Overview

Beard Meets Calculus reviews logarithm rules, repeated roots, and graph-based function composition before developing continuity through piecewise functions and continuous-function limit laws. The lecture then introduces the Intermediate Value Theorem and distinguishes horizontal asymptotes, found by studying inputs approaching ±∞, from vertical asymptotes, analyzed through one-sided limits and signs near a denominator zero.

Key takeaways

For nested limits such as g(f(x)), work from the innermost input outward and preserve one-sided direction: f(x) approaching 4 from below requires the left-hand behavior of f.
To make a piecewise function continuous, match the adjoining one-sided limits at every join; for the quadratic-line-quadratic example, the conditions a + b = 5 and −a + b = 3 yield a = 1 and b = 4.
When a complicated expression is built from continuous functions and its denominator is nonzero, evaluate its limit by direct substitution rather than manipulating every term.
The Intermediate Value Theorem guarantees every output between the endpoint values of a continuous function, but equal-sign endpoint values do not rule out roots inside the interval.
For (7e^x + 3)/(1 + e^x), the horizontal asymptotes differ by direction: y = 7 as x → +∞ and y = 3 as x → −∞.
For an infinite limit at a denominator zero, factor and check signs on both sides; x(x − 1)^2 stays positive near 1, making the example's two-sided limit +∞.

Chapters

0:00 Quickfire Review: Logarithm Rules and a Repeated Root

The logarithm of (x + 1)^4 divided by e^3(x − 4) separates into 4 ln(x + 1) − 3 − ln(x − 4).
Factoring x^3 − 2x^2 + x as x(x − 1)^2 reveals roots 0 and 1, with 1 counted twice.
Beard Meets Calculus encourages trying an answer—even zero or one—rather than avoiding participation.

3:55 Evaluate Nested Limits by Following Function Layers

For lim as x approaches 3 from below of g(f(x)), first track f(x): the graph shows it equals 2 nearby.
Then evaluate g(2) to get 1; avoid trying to construct the full composite function from the graphs.
For lim as x approaches 2 from above of f(g(x)), g(x) approaches 4 from below, so the relevant one-sided behavior of f gives 1.

8:35 Make a Piecewise Function Continuous at Its Join Points

The function uses 3x^2 for x < −1, ax + b between −1 and 1, and 4x^2 + 1 for x > 1.
Continuity at x = 1 requires a + b = 5, matching the middle line to the right-hand value of 4(1)^2 + 1.
Continuity at x = −1 requires −a + b = 3, matching the middle line to 3(−1)^2.
Solving the two equations gives a = 1 and b = 4; for piecewise functions, check continuity at each join.

20:58 Use Continuity to Evaluate Complicated Limits by Substitution

Polynomials, e^x, cosine, and arctangent are continuous; logarithm and tangent are continuous where they are defined and away from asymptotes.
Sums, products, quotients with nonzero denominators, and valid compositions of continuous functions remain continuous.
The displayed limit as x approaches 2 combines sine, exponential, cosine, a cubic–arctangent product, and a logarithm; continuity makes direct substitution sufficient.

24:55 Intermediate Value Theorem: Continuous Functions Cannot Skip Values

If f is continuous on an interval, it takes every value between f(a) and f(b) somewhere between a and b.
For 18x^3 − 63x^2 + 67x − 20, the values at 0 and 1 are −20 and 2, so continuity guarantees a root between 0 and 1.
The positive endpoint values f(1) = 2 and f(2) = 6 do not rule out a root between 1 and 2; the theorem guarantees intermediate values, not the absence of extra behavior.

30:30 Limits at Infinity and the Strategy for Horizontal Asymptotes

Infinity is not a number to substitute; limits as x approaches ±∞ describe inputs growing without bound in either direction.
A horizontal asymptote records a finite output approached as the input tends to positive or negative infinity.
For an ∞/∞ expression, identify the dominant growth term, factor it from numerator and denominator, and simplify; forms such as ∞ − ∞ may need rewriting as ratios.

34:30 Find Both Horizontal Asymptotes of an Exponential Ratio

For f(x) = (7e^x + 3)/(1 + e^x), the limit as x approaches +∞ is initially an ∞/∞ form.
Dividing numerator and denominator by e^x gives (7 + 3/e^x)/(1 + 1/e^x), which approaches 7; y = 7 is the positive-infinity horizontal asymptote.
As x approaches −∞, e^x approaches 0, so the original ratio approaches 3/1; y = 3 is the negative-infinity horizontal asymptote.

41:10 Vertical Asymptotes: Check One-Sided Behavior and Signs

A vertical asymptote occurs when the output grows without bound as x approaches a finite input, often when a nonzero numerator is divided by a denominator approaching zero.
Use one-sided limits because the denominator may approach zero from the positive or negative side, producing opposite signs of infinity.
To determine the direction of divergence, inspect the sign of the full expression near the denominator zero rather than considering only the numerator.

43:29 Sign Analysis Gives a Two-Sided Limit of Positive Infinity

For (3x^2 − 7x + 5)/(x^3 − 2x^2 + x), substitution at x = 1 gives a positive numerator, 1, over a zero denominator.
Factor the denominator as x(x − 1)^2; near 1, x is positive and (x − 1)^2 is positive on both sides.
The denominator approaches zero through positive values while the numerator stays positive, so the two-sided limit is +∞.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Beard Meets Calculus.