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What is the formal (ε–δ) definition of lim_{x→a} f(x) = L?
Answer: For every ε>0, there exists a δ>0 such that whenever 0 < |x−a| < δ, it follows that |f(x) − L| < ε.
This says f(x) can be made arbitrarily close to L (within ε) by taking x sufficiently close to a (within δ), without ever requiring x = a.
What is the difference between a left-hand limit lim_{x→a⁻} f(x) and a right-hand limit lim_{x→a⁺} f(x), and when does the two-sided limit exist?
Answer: The left-hand limit looks at values of f(x) as x approaches a from values less than a; the right-hand limit as x approaches a from values greater than a. lim_{x→a} f(x) exists if and only if both one-sided limits exist and are equal.
Example: for f(x)=|x|/x, lim_{x→0⁻}f(x)=−1 and lim_{x→0⁺}f(x)=1, so lim_{x→0}f(x) does not exist.
State the basic limit laws for sums, products, quotients, and constant multiples, given lim_{x→a} f(x)=L and lim_{x→a} g(x)=M.
Answer: lim[f(x)±g(x)] = L±M; lim[f(x)·g(x)] = L·M; lim[f(x)/g(x)] = L/M (if M≠0); lim[c·f(x)] = c·L.
Example: lim_{x→2}(3x^2+1) = 3·lim(x^2)+lim(1) = 3(4)+1 = 13.
If lim_{x→a} g(x) = b and f is continuous at b, what is lim_{x→a} f(g(x))?
Answer: lim_{x→a} f(g(x)) = f(b) = f(lim_{x→a} g(x)).
Example: if g(4)=9 and f(x)=√x is continuous at 9, then lim_{x→4}√(g(x)) = √9 = 3.
State the Squeeze (Sandwich) Theorem.
Answer: If g(x) ≤ f(x) ≤ h(x) for all x near a (except possibly at a), and lim_{x→a} g(x) = lim_{x→a} h(x) = L, then lim_{x→a} f(x) = L.
Classic example: since −x^2 ≤ x^2 sin(1/x) ≤ x^2 for all x≠0, and both bounds → 0 as x→0, the Squeeze Theorem gives lim_{x→0} x^2 sin(1/x) = 0.
How is a horizontal asymptote defined using limits?
Answer: y = L is a horizontal asymptote of f if lim_{x→∞} f(x) = L or lim_{x→−∞} f(x) = L.
Example: f(x) = (2x+1)/(x−3). As x→∞, f(x)→2, so y=2 is a horizontal asymptote.
For a rational function f(x)=P(x)/Q(x), how do you find lim_{x→±∞} f(x) by comparing the degrees of P and Q?
Answer: If deg P < deg Q, the limit is 0 (HA at y=0). If deg P = deg Q, the limit is the ratio of leading coefficients (HA at that value). If deg P > deg Q, the limit is ±∞ (no HA; a slant asymptote exists if deg P = deg Q + 1).
Example: f(x) = (2x^2+1)/(x^2+3) has equal degrees, so as x→±∞, f(x) → 2/1 = 2, giving horizontal asymptote y=2.
How is a vertical asymptote defined, and how do you locate it for a rational function?
Answer: x = c is a vertical asymptote if lim_{x→c⁻} f(x) = ±∞ or lim_{x→c⁺} f(x) = ±∞. For f(x)=P(x)/Q(x) in lowest terms, vertical asymptotes occur at the real zeros of Q(x).
Example: f(x) = 1/(x−2) has a vertical asymptote at x=2, where f(x)→+∞ as x→2⁺ and f(x)→−∞ as x→2⁻.
Evaluate lim_{x→∞} (3x^3 − 2x)/(5x^3 + 7).
Answer: 3/5
Divide numerator and denominator by the highest power x^3: limit = (3 − 2/x^2)/(5 + 7/x^3) → 3/5 as x→∞, consistent with the equal-degree rule.
What three conditions must all hold for f to be continuous at x = a?
Answer: (1) f(a) is defined, (2) lim_{x→a} f(x) exists, (3) lim_{x→a} f(x) = f(a).
If any one condition fails, f has a discontinuity at a.
Describe the three types of discontinuity: removable, jump, and infinite.
Answer: Removable: lim_{x→a} f(x) exists but ≠ f(a), or f(a) is undefined (a 'hole', fixable by redefining f(a)). Jump: the left-hand and right-hand limits both exist but are unequal. Infinite: f(x) → ±∞ as x→a (a vertical asymptote).
Example of a jump discontinuity: f(x)=x+1 for x<1 and f(x)=x−1 for x≥1. As x→1⁻, f→2; as x→1⁺, f→0 — the limits disagree, so there's a jump at x=1.
State the Intermediate Value Theorem (IVT).
Answer: If f is continuous on [a,b] and k is any value between f(a) and f(b), then there exists at least one c in [a,b] such that f(c) = k.
Example: f(x)=x^3−x−2 has f(1)=−2 and f(2)=4. Since 0 lies between −2 and 4 and f is continuous, IVT guarantees a root c in (1,2).
Write the limit definition of the derivative f'(x).
Answer: f'(x) = lim_{h→0} [f(x+h) − f(x)] / h
Example: for f(x)=x^2, f'(x) = lim_{h→0} [(x+h)^2 − x^2]/h = lim_{h→0} (2xh+h^2)/h = lim_{h→0}(2x+h) = 2x.
Give the alternate (two-point) form of the derivative at x = a.
Answer: f'(a) = lim_{x→a} [f(x) − f(a)] / (x − a)
This is algebraically equivalent to the difference-quotient form with x = a+h, and is especially useful for checking differentiability of piecewise functions at a specific point a.
What two interpretations does f'(a) have geometrically and physically?
Answer: f'(a) is the slope of the tangent line to y=f(x) at x=a, and it is the instantaneous rate of change of f with respect to x at x=a (e.g., velocity if f is a position function).
The tangent line at x=a has equation y − f(a) = f'(a)(x − a).
State the Power Rule and apply it to f(x) = x^5.
Answer: d/dx[x^n] = n·x^(n−1). For f(x)=x^5, f'(x) = 5x^4.
The rule works for any real exponent n, including negative and fractional powers, e.g., d/dx[x^(1/2)] = (1/2)x^(−1/2).
State the constant multiple rule and the sum/difference rule for derivatives.
Answer: d/dx[c·f(x)] = c·f'(x); d/dx[f(x) ± g(x)] = f'(x) ± g'(x).
Example: d/dx[3x^4 − 5x^2 + 7] = 12x^3 − 10x.
State the Product Rule and apply it to f(x) = x^2·sin x.
Answer: d/dx[u·v] = u'v + uv'. For f(x) = x^2 sin x: f'(x) = 2x sin x + x^2 cos x.
Here u=x^2 (so u'=2x) and v=sin x (so v'=cos x); substitute into u'v+uv'.
State the Quotient Rule and apply it to f(x) = sin x / x^2.
Answer: d/dx[u/v] = (u'v − uv')/v^2. For f(x) = sin x / x^2: f'(x) = (x cos x − 2 sin x)/x^3.
Here u=sin x (u'=cos x), v=x^2 (v'=2x): (cos x·x^2 − sin x·2x)/x^4 simplifies by factoring x to (x cos x − 2 sin x)/x^3.
State the Chain Rule and apply it to f(x) = sin(3x^2 + 1).
Answer: d/dx[f(g(x))] = f'(g(x))·g'(x). For f(x) = sin(3x^2+1): f'(x) = cos(3x^2+1)·6x.
Here the outer function is sin(u) and the inner function is u=3x^2+1, with u'=6x.
What are the derivatives of sin x, cos x, tan x, cot x, sec x, and csc x?
Answer: d/dx sin x = cos x; d/dx cos x = −sin x; d/dx tan x = sec²x; d/dx cot x = −csc²x; d/dx sec x = sec x tan x; d/dx csc x = −csc x cot x.
Worked example using the product rule: for f(x)=x tan x, f'(x) = tan x + x sec²x.
What is d/dx[e^x] and d/dx[a^x] for a > 0?
Answer: d/dx[e^x] = e^x; d/dx[a^x] = a^x·ln a.
Example: d/dx[2^x] = 2^x·ln 2. With the chain rule, d/dx[e^{3x}] = 3e^{3x}.
What is d/dx[ln x] and d/dx[log_a x]?
Answer: d/dx[ln x] = 1/x; d/dx[log_a x] = 1/(x ln a).
Example using the chain rule: d/dx[ln(x^2+1)] = 2x/(x^2+1).
What are the steps of implicit differentiation, and how do you find dy/dx for x² + y² = 25?
Answer: Differentiate both sides of the equation with respect to x, treating y as a function of x (applying the chain rule to y-terms, which produces a dy/dx factor), then solve algebraically for dy/dx. For x²+y²=25: 2x + 2y(dy/dx) = 0, so dy/dx = −x/y.
Implicit differentiation is used whenever y cannot be easily isolated as a function of x alone, such as for a circle equation.
If g is the inverse of f, what formula gives g'(x) in terms of f?
Answer: g'(x) = 1 / f'(g(x))
Example: f(x)=x^3+1 has f'(x)=3x^2. Since f(2)=9, the inverse g satisfies g(9)=2, so g'(9) = 1/f'(2) = 1/(3·4) = 1/12.
What are the derivatives of arcsin x, arccos x, and arctan x?
Answer: d/dx[arcsin x] = 1/√(1−x²); d/dx[arccos x] = −1/√(1−x²); d/dx[arctan x] = 1/(1+x²).
Example with the chain rule: d/dx[arctan(2x)] = 2/(1+4x²).
What do f''(x), f'''(x), and f^(n)(x) represent, and what do they mean for a position function s(t)?
Answer: They are the 2nd, 3rd, and nth derivatives of f, obtained by differentiating repeatedly. For position s(t): s'(t) is velocity, s''(t) is acceleration, and s'''(t) is jerk.
Example: s(t)=t^3−6t^2+9t gives velocity v(t)=3t^2−12t+9 and acceleration a(t)=6t−12.
Does differentiability imply continuity? Does continuity imply differentiability?
Answer: Differentiability at a point implies continuity there. The converse is false: a function can be continuous but not differentiable, such as f(x)=|x| at x=0.
f(x)=|x| is continuous everywhere, but it has a sharp corner at x=0 where the left-hand slope (−1) and right-hand slope (+1) disagree, so f'(0) does not exist.
List the ways a function can fail to be differentiable at a point.
Answer: (1) A corner, where left- and right-hand slopes differ; (2) a cusp, where slopes approach +∞ on one side and −∞ on the other; (3) a vertical tangent, where the slope approaches ±∞; (4) a discontinuity at that point.
Example: f(x)=x^(2/3) has a cusp at x=0; f(x)=∛x has a vertical tangent at x=0; both fail to be differentiable there despite being continuous.
State the Mean Value Theorem (MVT).
Answer: If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c in (a,b) such that f'(c) = [f(b) − f(a)] / (b − a).
Example: f(x)=x^2 on [0,2]. [f(2)−f(0)]/(2−0) = 4/2 = 2. Setting f'(c)=2c=2 gives c=1, which lies in (0,2).
State Rolle's Theorem and explain how it relates to the Mean Value Theorem.
Answer: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists some c in (a,b) with f'(c) = 0. Rolle's Theorem is the special case of the MVT when f(a) = f(b).
Example: f(x)=x^2−4x+3 on [1,3] has f(1)=f(3)=0. f'(x)=2x−4=0 gives c=2, which lies in (1,3).
State L'Hôpital's Rule and the indeterminate forms it applies to.
Answer: If lim_{x→a} f(x)/g(x) has indeterminate form 0/0 or ∞/∞, and f, g are differentiable near a with g'(x)≠0, then lim_{x→a} f(x)/g(x) = lim_{x→a} f'(x)/g'(x), provided the latter limit exists (or is ±∞).
L'Hôpital's Rule can be reapplied repeatedly if the new limit is still in indeterminate 0/0 or ∞/∞ form.
Evaluate lim_{x→0} (sin x)/x using L'Hôpital's Rule.
Answer: 1
This is a 0/0 form. Differentiating top and bottom gives lim_{x→0} cos x / 1 = cos(0) = 1, confirming the classic limit lim_{x→0} (sin x)/x = 1.
Can L'Hôpital's Rule be applied directly to forms like 0·∞ or ∞−∞? If not, how do you handle them?
Answer: No — L'Hôpital's Rule only applies directly to 0/0 or ∞/∞. Forms like 0·∞ or ∞−∞ must first be algebraically rewritten as a single quotient in 0/0 or ∞/∞ form before applying the rule.
Example: lim_{x→0⁺} x ln x (a 0·∞ form) is rewritten as lim_{x→0⁺} ln x/(1/x) (∞/∞ form). Applying L'Hôpital: lim (1/x)/(−1/x²) = lim(−x) = 0.
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