What does lim_{x→a} f(x) = L mean informally? — f(x) gets close to L as x gets close to a.The limit L means f(x) can be made arbitrarily close to L by taking x sufficiently close to a — without requiring f(a) = L.
Evaluate lim_{x→4} (x + 2) by direct substitution. — 6.Direct substitution: 4 + 2 = 6. Polynomial functions are continuous everywhere, so substitution works.
Does the function need to be defined at x = a for lim_{x→a} f(x) to exist? — No, the limit is independent of f(a).The limit describes the behavior of f(x) as x approaches a, not the value at a. The limit can exist even if f(a) is undefined.
Evaluate lim_{x→0} 5 (a constant function). — 5.The limit of a constant is that constant: lim_{x→a} c = c for any value of a.
What is lim_{x→2} x²? — 4.Direct substitution: 2² = 4. The limit of xⁿ as x→a is aⁿ for positive integers n.
If lim_{x→a} f(x) = 3 and lim_{x→a} g(x) = 5, what is lim_{x→a} [f(x) + g(x)]? — 8.By the sum limit law: lim[f+g] = lim f + lim g = 3 + 5 = 8.
What indeterminate form signals the need for algebraic manipulation? — 0/0.0/0 is an indeterminate form — direct substitution fails, and we need to factor, cancel, or rationalize to find the actual limit.
The left-hand limit lim_{x→a⁻} f(x) considers x approaching a from: — Values less than a.The left-hand (or left-sided) limit uses x values approaching a from the left, i.e., x < a.
For the two-sided limit lim_{x→a} f(x) to exist, what must be true? — The left and right limits must both exist and be equal.The two-sided limit exists if and only if both one-sided limits exist AND are equal to each other.
What is lim_{x→∞} (1/x)? — 0.As x grows without bound, 1/x approaches 0. This is a fundamental limit at infinity.
What does lim_{x→∞} f(x) = L mean geometrically? — The graph of f approaches the horizontal line y = L as x → ∞.The horizontal asymptote y = L is the geometric interpretation of lim_{x→∞} f(x) = L.
lim_{x→∞} (5/x³) equals: — 0.As x→∞, 5/x³ → 0 since the denominator grows much faster than the constant numerator.
For f(x) = 1/(x−3), what is lim_{x→3⁺} f(x)? — +∞.As x→3⁺, (x−3)→0⁺, so 1/(x−3)→+∞.
What is the horizontal asymptote of f(x) = (3x + 1)/(x − 2)? — y = 3.lim_{x→∞}(3x+1)/(x−2) = lim(3 + 1/x)/(1 − 2/x) = 3/1 = 3. Horizontal asymptote: y = 3.
For f(x) = {x² if x < 1; 2 if x ≥ 1}, find lim_{x→1⁻} f(x). — 1.For x < 1, f(x) = x². As x→1⁻, x²→1² = 1. Left-hand limit = 1.
A function f is continuous at x = a if: — All three: f(a) defined, limit exists, and they are equal.Continuity at x = a requires: (1) f(a) is defined, (2) lim_{x→a} f(x) exists, and (3) lim_{x→a} f(x) = f(a).
Which type of function is always continuous on its entire domain? — Polynomial functions.Polynomial functions are continuous everywhere on ℝ — they have no breaks, holes, or asymptotes.
A "hole" in the graph of a function represents: — Removable discontinuity.A removable discontinuity appears as a hole — the limit exists, but the function value is either missing or differs from the limit.
What is the Intermediate Value Theorem (IVT) used for? — Proving the existence of a zero between two points.The IVT guarantees that if a continuous function changes sign on [a,b], then it has at least one zero (root) in (a,b).
Is f(x) = 1/x continuous at x = 0? — No, it has an infinite discontinuity.At x = 0, f is undefined and lim_{x→0} 1/x = ±∞. This is an infinite discontinuity (vertical asymptote).
If f is continuous on [a, b] and f(a) = −3 and f(b) = 5, does f have a root in (a, b)? — Yes, by the IVT since f changes sign.By the IVT, since f is continuous and changes from negative to positive, there exists c ∈ (a,b) with f(c) = 0.
A jump discontinuity occurs when: — The left and right limits exist but are not equal.In a jump discontinuity, both one-sided limits exist as finite numbers but are unequal. The graph has a step or jump.
Is f(x) = |x| continuous at x = 0? — Yes, it is continuous.f(x) = |x| is continuous at 0: f(0) = 0, lim_{x→0}|x| = 0, they agree. (It is not differentiable at 0, but it IS continuous.)
What are the two main indeterminate forms that L'Hôpital's Rule directly handles? — 0/0 and ∞/∞.L'Hôpital's Rule applies directly when a limit yields the indeterminate forms 0/0 or ∞/∞.
In L'Hôpital's Rule, you differentiate: — The numerator and denominator separately.L'Hôpital's Rule: lim f/g = lim f'/g'. You differentiate f and g separately — NOT using the quotient rule.
Evaluate lim_{x→0} sin(x)/x. — 1.0/0 form. L'Hôpital: lim cos(x)/1 = 1. This is the fundamental trigonometric limit.
What is the indeterminate form of lim_{x→∞} (eˣ/x²)? — ∞/∞.As x→∞, both eˣ and x² grow without bound, giving ∞/∞ — an indeterminate form.
Can L'Hôpital's Rule be applied to lim_{x→2} (x²−4)/(x−2)? — Yes, and the result is 4.At x=2: 0/0 form. L'Hôpital: lim (2x)/1 = 4. (Factoring also gives 4.)
What is lim_{x→0} (eˣ − 1)/x? — 1.0/0 form. L'Hôpital: lim eˣ/1 = e⁰ = 1. This is the definition of the derivative of eˣ at 0.
Before applying L'Hôpital's Rule, you must always verify: — The limit gives an indeterminate form.L'Hôpital's Rule is only valid for indeterminate forms (0/0, ∞/∞). Applying it to non-indeterminate limits gives wrong answers.