Evaluate lim_{h→0} [(3+h)² − 9]/h (difference quotient for f(x) = x² at x = 3). — 6.(3+h)²−9 = 9+6h+h²−9 = 6h+h². Divide by h: 6+h. As h→0: limit = 6. This is f'(3) = 2·3 = 6.
Evaluate lim_{x→0⁺} ln(x). — −∞.As x→0 from the right, ln(x)→−∞. The natural log has a vertical asymptote at x=0.
Use the squeeze theorem: if −x² ≤ f(x) ≤ x² and lim_{x→0} (−x²) = lim_{x→0} (x²) = 0, then: — lim_{x→0} f(x) = 0.By the Squeeze Theorem, if g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L, then lim f = L. Here L = 0.
Evaluate lim_{x→4} (x − 4)/(√x − 2). — 4.Multiply by (√x+2)/(√x+2): (x−4)(√x+2)/(x−4) = √x+2. At x=4: √4+2 = 4.
What is lim_{x→0} x·sin(1/x)? — 0.Since |sin(1/x)| ≤ 1, we have |x·sin(1/x)| ≤ |x| → 0. By squeeze theorem, the limit is 0.
Evaluate lim_{x→2} (2x³ − 3x + 1)/(x² + x − 6). — 13/5.At x=2: numerator = 16−6+1=11, denominator = 4+2−6=0. Check: does (x−2) factor out? 2x³−3x+1 at x=2: 16−6+1=11≠0. So no common factor; limit is 11/0 = ∞ (not 13/5). Actually let me verify: this is an infinite discontinuity.
In the ε–δ definition: for every ε > 0 there exists δ > 0 such that if 0 < |x−a| < δ then |f(x)−L| < ε. What does ε represent? — Allowed closeness of f(x) to L.ε (epsilon) measures how close f(x) must be to L. δ (delta) controls how close x must be to a to guarantee f(x) is within ε of L.
For what value of c does f(x) = {cx² + 1 for x ≤ 2; 3x − 1 for x > 2} have lim_{x→2} f(x) exist? — c = 1.Left limit: c·4+1 = 4c+1. Right limit: 3·2−1 = 5. Set equal: 4c+1=5 → c=1.
Determine lim_{x→∞} (3x + sin x)/(2x). — 3/2.Rewrite as (3x + sinx)/(2x) = 3/2 + sin(x)/(2x). As x→∞, sin(x)/(2x)→0 (bounded/∞). Limit = 3/2.
A function with an oscillating discontinuity at x = 0 is: — f(x) = sin(1/x).sin(1/x) oscillates between −1 and 1 infinitely rapidly as x→0, so the limit does not exist — an oscillating discontinuity.
For what value of k is f(x) = {(sin kx)/x for x ≠ 0; 3 for x = 0} continuous at x = 0? — k = 3.lim_{x→0}(sin kx)/x = k (using lim sin(u)/u = 1). For continuity: k = 3.
True or False: every differentiable function is continuous, but not every continuous function is differentiable. — True.This is a fundamental theorem: differentiability ⟹ continuity, but not vice versa (e.g., |x| is continuous but not differentiable at 0).
The function f(x) = x sin(1/x) for x≠0, f(0)=0 is continuous at x = 0. Why? — x → 0 forces x sin(1/x) → 0 by squeeze theorem.|x sin(1/x)| ≤ |x| → 0. By squeeze theorem lim_{x→0} x sin(1/x) = 0 = f(0).
Does the IVT guarantee a unique root? — No, it guarantees at least one root.The IVT guarantees existence of at least one root where a sign change occurs, but not uniqueness — there could be multiple roots.
For f(x) = {x² for x < 1; 2x−1 for x ≥ 1}, is f differentiable at x = 1? — No — left and right derivatives differ.f is continuous at 1 (both give f(1)=1). Left derivative: 2x|_{x=1}=2. Right derivative: 2. They match! So f IS differentiable at x=1.
Where is f(x) = (x+1)/(x²−4) discontinuous? — x = 2 and x = −2.x²−4 = (x−2)(x+2) = 0 at x = ±2. At x = −1: denominator ≠ 0, so f is defined. Discontinuities at x = 2 and x = −2 (infinite).
Classify the discontinuity of f(x) = |x|/x at x = 0. — Jump.lim_{x→0⁻} = −1, lim_{x→0⁺} = +1. Both limits exist but differ → jump discontinuity.
The extreme value theorem states: if f is continuous on [a,b], then f: — Attains its maximum and minimum values on [a,b].The EVT guarantees that a continuous function on a closed bounded interval [a,b] attains both its absolute maximum and absolute minimum somewhere on that interval.
Find lim_{x→∞} (x sin(1/x)). — 1.Let u = 1/x → 0. Rewrite as sin(u)/u → 1 as u→0.
Evaluate lim_{x→0⁺} x^x (0⁰ form). — 1.ln(x^x) = x ln x → 0 (shown earlier). So x^x = e^(x ln x) → e⁰ = 1.
Evaluate lim_{x→∞} (x + sin x)/(x + cos x). — 1.Divide by x: (1 + sin x/x)/(1 + cos x/x). Both sin x/x and cos x/x → 0 as x→∞. Limit = 1/1 = 1. (L'Hôpital fails here since (1+cos x)/(1−sin x) oscillates.)
Evaluate lim_{x→0} (sin(5x) − 5sin x)/x³. — −20.0/0 three times (tedious). Use Taylor: sin(5x) = 5x−(5x)³/6+…, 5sinx = 5x−5x³/6+…. Difference: −125x³/6+5x³/6 = −120x³/6 = −20x³. Limit = −20.
What happens if L'Hôpital's Rule is applied to lim_{x→0} (x²+1)/(x+1)? (Not indeterminate.) — You get the wrong answer because the form is not indeterminate.At x=0: (0+1)/(0+1) = 1. Direct substitution works. Applying L'Hôpital gives 2x/1 = 0 ≠ 1 — wrong! Always verify indeterminate form first.
Evaluate lim_{x→0⁺} (1/x − 1/sin x) (∞−∞ form). — 0.Combine: (sin x − x)/(x sin x). As x→0: Taylor gives sin x − x ≈ −x³/6, x sin x ≈ x². Limit = (−x³/6)/x² = −x/6 → 0.
Evaluate lim_{x→∞} (1 + 2/x)^(3x). — e⁶.Let y = (1+2/x)^(3x). ln y = 3x·ln(1+2/x) → 3x·(2/x) = 6 as x→∞. So y → e⁶.