Evaluate lim_{x→0} (sin x)/x. — 1.This is a fundamental trigonometric limit: lim_{x→0} sin(x)/x = 1. It can be proven geometrically or using the squeeze theorem.
If lim f(x) = 4 and lim g(x) = 2, what is lim [f(x)·g(x)]? — 8.Product law: lim[f·g] = lim f · lim g = 4·2 = 8.
Evaluate lim_{x→9} (√x − 3)/(x − 9) by rationalizing. — 1/6.Multiply by (√x+3)/(√x+3): (x−9)/[(x−9)(√x+3)] = 1/(√x+3). At x=9: 1/6.
Does lim_{x→0} |x|/x exist? — No, it does not exist.From the right: |x|/x = x/x = 1. From the left: |x|/x = −x/x = −1. Since one-sided limits differ, the two-sided limit does not exist.
Evaluate lim_{x→∞} 5 (constant). — 5.A constant function always has value 5, regardless of x. Even as x→∞, the limit is 5.
What is lim_{x→0} (1 − cos x)/x²? — 1/2.This is a standard limit: lim_{x→0}(1−cosx)/x² = 1/2. Can be derived using Taylor series or L'Hôpital's rule.
What does it mean for a limit to fail to exist due to oscillation? — The function alternates between values without settling.Oscillating limits (like sin(1/x) at x=0) never settle on a specific value — the function oscillates indefinitely, so no limit exists.
Does lim_{x→1} f(x) exist for f(x) = {x² if x < 1; 2 if x ≥ 1}? — No, left and right limits differ.Left limit = 1 (from x²), right limit = 2 (constant). Since 1 ≠ 2, the two-sided limit does not exist.
lim_{x→∞} (7x³ − 2x)/(3x³ + x²) equals: — 7/3.Divide by x³: (7 − 2/x²)/(3 + 1/x) → 7/3 as x→∞.
lim_{x→∞} (2x² + 5)/(x³ − 1) equals: — 0.Numerator degree (2) < denominator degree (3). The limit is 0.
lim_{x→∞} (x³ + x)/(x² + 1) equals: — ∞.Numerator degree (3) > denominator degree (2). The limit is +∞ (grows without bound).
For the rational function f(x) = (2x+3)/(5x−1), what is lim_{x→−∞} f(x)? — 2/5.For rational functions with equal degrees, lim at ±∞ equals the ratio of leading coefficients: 2/5. The sign of x doesn't change the leading-coefficient result.
At x = 0, f(x) = 1/x² has lim_{x→0} f(x) equal to: — +∞.1/x² is always positive and grows without bound as x→0 from either side. lim_{x→0} 1/x² = +∞.
lim_{x→∞} sin(x) equals: — Does not exist.sin(x) oscillates between −1 and 1 forever as x→∞. It never settles on a value, so the limit does not exist.
What is lim_{x→0⁻} (1/x)? — −∞.As x→0 from the left, x is a small negative number. 1/x becomes a large negative number → −∞.
For the floor function f(x) = ⌊x⌋, what is lim_{x→2} f(x)? — Does not exist.From the left: ⌊x⌋=1 as x→2⁻. From the right: ⌊x⌋=2 as x→2⁺. One-sided limits differ, so the two-sided limit does not exist.
For f(x) = (x² − x − 6)/(x − 3), what type of discontinuity is at x = 3? — Removable.x²−x−6 = (x−3)(x+2). Cancel: f = x+2 for x≠3. Limit = 5 but f(3) undefined. Removable discontinuity.
f is continuous on [1, 5] with f(1) = 4 and f(5) = −2. By IVT, what value must f hit in (1, 5)? — 0 and any value between −2 and 4.By IVT, every value k with −2 < k < 4 is achieved by f at some c ∈ (1, 5), including k = 0.
Find the value of c that makes f(x) = {3x + c for x < 2; x² − 1 for x ≥ 2} continuous at x = 2. — c = −1.For continuity: left limit = right limit. 3(2)+c = 2²−1 → 6+c = 3 → c = −3.
Does g(x) = tan(x) have any discontinuities on [0, 2π]? — Yes, at x = π/2 and x = 3π/2.tan(x) = sin(x)/cos(x). It is undefined where cos(x) = 0, i.e., at x = π/2 and x = 3π/2. These are infinite discontinuities.
Apply the IVT: show f(x) = x³ − 2x − 5 has a root in (2, 3). What are f(2) and f(3)? — f(2) = −1, f(3) = 16.f(2) = 8−4−5 = −1 < 0. f(3) = 27−6−5 = 16 > 0. Sign change on continuous function → root exists in (2,3).
Is f(x) = sin(1/x) continuous at x = 0? — No — oscillating discontinuity (limit does not exist).sin(1/x) oscillates infinitely as x→0. The limit does not exist (oscillating discontinuity), so f is not continuous at 0.
If f is not continuous at x = a, can f still be differentiable at x = a? — No — differentiability requires continuity.Differentiability implies continuity. A function cannot be differentiable at a point where it is discontinuous.
Which is an example of a function continuous everywhere on ℝ? — f(x) = eˣ.eˣ is continuous for all real x. 1/(x²−1) has discontinuities at ±1. √x is only defined for x ≥ 0. ln x requires x > 0.
Evaluate lim_{x→∞} (ln x)/x. — 0.∞/∞ form. L'Hôpital: lim (1/x)/1 = 0. Logarithm grows much slower than x.
Evaluate lim_{x→0} (1 − cos x)/x². — 1/2.0/0 form. First derivative: sin x / 2x → 0/0. Second derivative: cos x / 2 = 1/2.
To handle lim_{x→0⁺} x ln(x) (0·∞ form), rewrite as: — ln(x)/(1/x).Write x·ln x = ln x / (1/x) to get −∞/+∞ form, then apply L'Hôpital.
Evaluate lim_{x→0⁺} x ln(x). — 0.Rewrite as ln x/(1/x) → −∞/+∞. L'Hôpital: (1/x)/(−1/x²) = −x → 0.
For 1^∞ forms like lim_{x→∞}(1+1/x)^x, the first step is to: — Take the natural log: let y = (1+1/x)^x, find lim ln y.For 1^∞ forms, let y = f(x)^g(x), then ln y = g(x)·ln f(x). Find lim ln y (now 0·∞ or 0/0), then the answer is e^(that limit).