More limits (Calc 1; Lecture 1-3; Fall 26)

Overview

Beard Meets Calculus develops a practical strategy for evaluating limits: substitute first, then use algebraic rewrites—factoring, conjugates, or trigonometric identities—to remove indeterminate 0/0 forms. The lecture proves the squeeze principle and the radians-based limit sin(θ)/θ → 1, then applies that result to limits involving sine.

Key takeaways

Direct substitution is the first test for a limit: a 0/0 result means the form is indeterminate and requires rewriting, not that the limit equals zero.
Factor common terms and cancel them only for nearby inputs where the original expression is defined; the simplified expression can then reveal the limit.
Multiplying by a conjugate converts radical differences into differences of squares, often exposing a factor that cancels the zero-producing denominator term.
The squeeze principle evaluates t sin(1/t) at zero despite infinitely rapid oscillation, because its magnitude is bounded above by |t|.
The foundational limit sin θ/θ → 1 depends on measuring θ in radians and supports approximating sin θ by θ for small angles.
For scaled-angle limits, match the denominator to the sine’s inside expression—for example, rewrite sin(2θ)/θ as 2·sin(2θ)/(2θ)—rather than incorrectly treating sin(2θ) as 2 sin θ.

Chapters

0:00 Quickfire Review: Exponent Rules and Function Substitution

Combining e^(5t)(e^(4t))^3/e^(-6t) gives e^(23t): multiply exponents in the power, then subtracting a negative adds 6t.
For f(x) = x² + 4x, evaluating f(2 + h) − f(2) gives h² + 8h.
The substitution example previews a recurring Calc 1 skill: use parentheses when plugging expressions into a function.

3:09 Evaluating a Difference of Rational Expressions at x = 3

Substituting x = 3 into both fractions produces nonzero-over-zero expressions, so the difference is indeterminate and needs algebraic simplification.
Since the fractions share denominator x − 3, combine their numerators as (x² − x) − (9 − x), carefully distributing the minus sign.
Factor x² − 9 as (x − 3)(x + 3), cancel x − 3 for x near—but not equal to—3, and obtain the limit 6.

7:20 Limit Strategy: Work Through Frustration and Look for Cancellation

Steve frames frustration as a normal part of learning mathematics and encourages students to keep working until concepts click.
For a limit that substitutes to 0/0, treat the form as a signal for more work rather than as an answer.
Useful rewrites include factoring, expanding, multiplying by a conjugate, combining fractions, and applying trigonometric identities.

11:20 Factoring a 0/0 Limit to Get the Value 7

For the limit as x approaches 2 of (x² + 3x − 10)/((x − 3)² + 3x − 7), direct substitution gives 0/0.
The numerator factors as (x − 2)(x + 5); expanding and combining the denominator gives x² − 3x + 2 = (x − 2)(x − 1).
Cancel the shared x − 2 factor, then substitute x = 2 into (x + 5)/(x − 1) to get 7.

15:45 Using a Conjugate to Evaluate a Radical Limit

For the limit as t approaches 4 of (t − 4)/(√(t² − 7) − 3), substitution produces 0/0.
Multiply numerator and denominator by the conjugate √(t² − 7) + 3, preserving the expression by multiplying by an equivalent form of 1.
The denominator becomes (t² − 7) − 9 = t² − 16 = (t − 4)(t + 4); cancel t − 4 and evaluate to get 6/8 = 3/4.
Expand only where it helps: leave the numerator’s product intact so its t − 4 factor remains easy to cancel.

22:05 Trig Identity Simplification Gives a Limit of One-Half

For (1 − cos θ)/sin² θ as θ approaches 0, substitution gives 0/0.
Use sin² θ = 1 − cos² θ, then factor the denominator as (1 − cos θ)(1 + cos θ).
Cancel the common factor 1 − cos θ and substitute θ = 0 into 1/(1 + cos θ), yielding 1/2.
Cancellation applies to shared factors, not visually similar pieces of sums or differences.

26:44 The Squeeze Principle and the Limit of t sin(1/t)

If g(x) ≤ f(x) ≤ h(x), and g and h approach the same value L as x approaches c, the squeeze principle forces f to approach L too.
Although sin(1/t) oscillates infinitely often near t = 0, it remains bounded between −1 and 1.
Multiplying those bounds by the magnitude of t gives −|t| ≤ t sin(1/t) ≤ |t|; both bounds approach 0, so t sin(1/t) approaches 0.

32:17 Unit-Circle Areas Prove sin(θ)/θ Approaches 1

For 0 < θ < π/2 on the unit circle, compare the areas of an inscribed triangle, a circular sector, and a tangent-based triangle.
Their areas are respectively (1/2)sin θ, (1/2)θ, and (1/2)tan θ, giving sin θ < θ < tan θ.
Rearranging yields cos θ < sin θ/θ < 1; the argument uses radians because the sector’s area is θ/2.
Even symmetry extends the bounds to negative θ near zero; since both outer expressions approach 1, the squeeze principle proves lim(θ→0) sin θ/θ = 1.

40:48 Applying sin(θ)/θ to Limits with Scaled Angles

The limit sin θ/θ = 1 implies sin θ is approximately θ for small angles, with θ measured in radians.
For sin(2θ)/θ, either use sin(2θ) = 2 sin θ cos θ or rewrite it as 2·[sin(2θ)/(2θ)]; both methods give 2.
More generally, sin(u)/u approaches 1 whenever u approaches 0, allowing the inside expression to serve as the matching denominator.
For sin(9t²)/(t sin(3t)) as t approaches 0, use sin(9t²) ≈ 9t² and sin(3t) ≈ 3t to obtain the limit 3.

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.