Limits, L'Hôpital's Rule & Mean Value Theorems
“Indeterminate forms 0/0, inf/inf, 1^inf, L'Hôpital's rule, Rolle's theorem, and Lagrange's MVT average slope formula.”
1.Limits, Indeterminate Forms & Mean Value Theorems
§ 1.1.Limits describe the behavior of functions near points where direct evaluation fails or yields singular divisions.
L'Hôpital's Rule:
Applies directly to indeterminate forms 0/0 and infinity/infinity:
lim_{x -> a} f(x) / g(x) = lim_{x -> a} f'(x) / g'(x).
For exponential forms 1^infinity, 0^0, and infinity^0, we transform via logarithms: ln y = g(x) ln f(x) = ln f(x) / (1 / g(x)) to reduce to 0/0.
Mean Value Theorems:
- Rolle's Theorem: If f(a) = f(b) on a differentiable interval, the tangent must be horizontal (f'(c) = 0) at least once.
- Lagrange's MVT: Connects secant line slope to instantaneous slope: f'(c) = (f(b) - f(a)) / (b - a).
In machine learning, MVT is the foundation for proving Lipschitz continuity and bounding gradient growth.
2. Worked Practice Problem: Exponential Limit & MVT Root Interval
Use the logarithmic transformation ln(y) to resolve the 1^inf indeterminate form.
Problem: (a) Evaluate L = lim_{x -> 0} (1 + sin x)^{1/x}. (b) For f(x) = x^3 - x on [0, 2], find the value of c in (0, 2) satisfying Lagrange's Mean Value Theorem.
Part (a): Evaluate L = lim_{x -> 0} (1 + sin x)^{1/x}:
Step 1: Check form at x = 0: (1 + 0)^{1/0} = 1^infinity (Indeterminate form).
Step 2: Let y = (1 + sin x)^{1/x} => ln y = (1/x) * ln(1 + sin x) = ln(1 + sin x) / x.
Step 3: As x -> 0, ln(1 + 0) / 0 = 0/0. Apply L'Hôpital's rule:
Numerator derivative: d/dx [ln(1 + sin x)] = cos x / (1 + sin x).
Denominator derivative: d/dx [x] = 1.
lim_{x -> 0} [ cos x / (1 + sin x) ] / 1 = cos(0) / (1 + sin 0) = 1 / 1 = 1.
Step 4: Exponentiate to recover L: ln L = 1 => L = e^1 = e ≈ 2.71828.
Part (b): Find c for f(x) = x^3 - x on [0, 2]:
Step 1: Compute average slope: f(0) = 0; f(2) = 2^3 - 2 = 8 - 2 = 6.
Average slope = [ f(2) - f(0) ] / [ 2 - 0 ] = 6 / 2 = 3.
Step 2: Set instantaneous derivative f'(c) equal to 3:
f'(x) = 3*x^2 - 1.
3c^2 - 1 = 3 => 3c^2 = 4 => c^2 = 4 / 3.
c = 2 / sqrt(3) ≈ 1.1547 (since c must lie in the open interval (0, 2)).
Conclusion: (a) L = e; (b) c = 2 / sqrt(3) ≈ 1.1547.
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