Day 17: Limits, derivatives, and nondifferentiability

(a) f(3 + h) = (3 + h)^2 = 9 + 6h + h^2.
f(3 + h) - f(3) = 9 + 6h + h^2 - 9 = 6h + h^2.
[f(3 + h) - f(3)]/h = (6h + h^2)/h = 6 + h, for h != 0 (the cancellation divides by h).
lim(h->0) (6 + h) = 6, so f'(3) = 6. Both one-sided limits are 6, because 6 + h approaches 6 whether h is positive or negative.
Numerical check: h = 0.01 gives (9.0601 - 9)/0.01 = 6.01; h = -0.01 gives (8.9401 - 9)/(-0.01) = 5.99. Both are 0.01 away from 6, as 6 + h predicts.

(b) [(a + h)^2 - a^2]/h = (2ah + h^2)/h = 2a + h for h != 0, so f'(a) = 2a. At a = 3 this gives 6 again; at a = 2 it gives 4 (the mastery item). This is the power rule for n = 2, derived rather than assumed.

(c) g(2 + h) - g(2) = 1/(2 + h) - 1/2 = (2 - (2 + h))/(2(2 + h)) = -h/(2(2 + h)).
Divide by h (h != 0, and h != -2 so that 2 + h is not zero): -1/(2(2 + h)).
As h approaches 0 this tends to -1/(2 * 2) = -1/4, so g'(2) = -0.25.
Numerical check: h = 0.01 gives (1/2.01 - 0.5)/0.01 = (0.497512 - 0.5)/0.01 ≈ -0.2488, close to -0.25.

Common errors: writing (3 + h)^2 = 9 + h^2 (the 6h is missing); setting h = 0 before cancelling, which gives 0/0; leaving the answer as 6 + h.

(a) Term by term:
d/dx x^3 = 3x^2 (power rule, n = 3)
d/dx (-2x) = -2 (constant multiple of the power rule with n = 1: -2 * 1 * x^0)
d/dx 1 = 0 (constant rule)
Sum rule: q'(x) = 3x^2 - 2 + 0 = 3x^2 - 2.
At x = 2: q'(2) = 3(2)^2 - 2 = 3 * 4 - 2 = 12 - 2 = 10.
Central-difference check with h = 0.001: q(2.001) = 8.012006001 - 4.002 + 1 = 5.010006001 and q(1.999) = 7.988005999 - 3.998 + 1 = 4.990005999. (5.010006001 - 4.990005999)/0.002 = 0.020000002/0.002 = 10.000001. The extra 0.000001 is exactly h^2 for this cubic, the truncation term of the central difference; it is not an error in q'(2).

(b) 3x^2 - 2 = 0 gives x^2 = 2/3, so x = sqrt(2/3) ≈ 0.8165 or x = -sqrt(2/3) ≈ -0.8165. At these two points the tangent line is horizontal.

(c) u'(x) = 10x^4 - 6x + 7, so u'(1) = 10 - 6 + 7 = 11.

Domain note: polynomials are defined and differentiable for every real x, so no point needs to be excluded. The rules are shortcuts for the limit definition, and each one is proved from it.

Common errors: differentiating the exponent (writing 3x^3 or x^2); keeping the constant 1 as 1; substituting x = 2 before differentiating.

(a) Continuity: lim(x->0) |x| = 0 = v(0), so v is continuous at 0.
Quotient: [v(0 + h) - v(0)]/h = |h|/h.
For positive h, |h| = h, so the quotient is h/h = 1; at h = 0.1 it is 0.1/0.1 = 1.
For negative h, |h| = -h, so the quotient is -h/h = -1; at h = -0.1 it is 0.1/(-0.1) = -1.
The right-hand limit is 1 and the left-hand limit is -1. They differ, so the two-sided limit does not exist and v'(0) does not exist, even though v is continuous at 0. This is a counterexample to "continuous always means differentiable." Geometrically, lines of slope 1 and -1 meet at a corner at the origin.

(b) w(0) = 0, so the quotient is w(h)/h = h|h|/h = |h| for h != 0. As h approaches 0 from either side, |h| approaches 0. Both one-sided limits are 0, so w'(0) = 0. Multiplying by x removed the corner: the graph is -x^2 on the left and x^2 on the right, and they meet with the same slope 0.

(c) c(h)/h = h^(1/3)/h = h^(-2/3). At h = 0.001: 0.1/0.001 = 100. At h = -0.001: (-0.1)/(-0.001) = 100. As h approaches 0 the quotient grows without bound (h^(-2/3) tends to +infinity), so c has a vertical tangent at 0 and no finite derivative, although it is continuous there.

Summary: corner (|x|, unequal one-sided limits), smooth join (x|x|, equal limits), vertical tangent (cube root, infinite limit). A function with a jump is never differentiable at the jump, because differentiability implies continuity.

Rules: s'(x) = 2x - 4 (power rule on x^2, constant multiple on -4x). So s'(3) = 2(3) - 4 = 2.
Limit-definition check: s(3 + h) = (3 + h)^2 - 4(3 + h) = 9 + 6h + h^2 - 12 - 4h = -3 + 2h + h^2. With s(3) = -3, [s(3 + h) - s(3)]/h = (2h + h^2)/h = 2 + h for h != 0, which tends to 2. Both methods agree.

Tangent line: s(3) = 9 - 12 = -3, so the point is (3, -3) and the slope is 2: y + 3 = 2(x - 3), or y = 2x - 9. Check at x = 3: 2(3) - 9 = -3.

Local linear approximation: at x = 3.1 the tangent gives 2(3.1) - 9 = -2.8; the true value is s(3.1) = 9.61 - 12.4 = -2.79. The error 0.01 is h^2 for h = 0.1, the term the line ignores.

Contrast with |x| at 0: unlike the smooth polynomial, |x| has unequal one-sided slopes at 0 and no derivative there: explicitly, lim(h->0+) (|h| - |0|)/h = +1 while lim(h->0-) (|h| - |0|)/h = -1. A polynomial has a tangent line at every point; |x| has none at 0, so no single line approximates its corner from both sides.

Variant for the check: s'(1) = -2 and s(1) = -3, so the tangent at x = 1 is y = -2x - 1.

Common errors: giving the slope 2 as the whole answer; using s(3) = 3; giving |x| the derivative 0 at 0 by averaging +1 and -1.