The formal definition of a limit is a cornerstone of calculus. It gives a precise meaning to the statement "the limit of f(x) as x approaches 'a' is L".
Definition: We saylim (x→a) f(x) = Lif for any given numberε > 0(no matter how small), there exists a corresponding numberδ > 0such that:
|f(x) - L| < εwhenever0 < |x - a| < δ
|f(x) - L| < ε means "the distance between f(x) and L is less than ε". This is the output tolerance.|x - a| < δ means "the distance between x and 'a' is less than δ". This is the input tolerance.0 < ... part just means that x ≠ a. We don't care what happens *at* x = a, only *near* it.In simple terms: "You can make f(x) as close as you want (ε) to L, just by making x close enough (δ) to 'a'."
This is a theoretical tool that lets us prove a limit *exists* without knowing what the limit's value (L) is.
Statement (without proof): A function f(x) has a finite limit as x approaches 'a' if and only if for every ε > 0, there exists a δ > 0 such that:
|f(x₁) - f(x₂)| < ε whenever 0 < |x₁ - a| < δ and 0 < |x₂ - a| < δ.
In simple terms: "If x is close to 'a', the function's values f(x) must be close to each other."
Continuity is a more strict condition than just having a limit. A function is continuous at a point if the limit *exists*, the function is *defined* at that point, and the *limit equals the function's value*.
Definition: A function f(x) is continuous at a pointx = aif for anyε > 0, there exists aδ > 0such that:
|f(x) - f(a)| < εwhenever|x - a| < δ
Note the difference from the limit definition:
f(x) to f(a), not some limit L.0 < |x - a|. This means we *include* the point x = a itself, which is fine since |f(a) - f(a)| = 0, which is always less than ε.If a function is not continuous at x = a, it is discontinuous. There are several types:
lim (x→a) f(x) = L exists.f(a) is not defined, *or* f(a) ≠ L.f(x) = (x² - 1) / (x - 1) at x = 1. The limit is 2, but f(1) is undefined. It's a "hole" in the graph that can be "removed" by defining f(1) = 2.lim (x→a⁻) f(x) and the right-hand limit lim (x→a⁺) f(x) both exist, but they are *not equal*.f(x) = [x] at x = 2. The left limit is 1, and the right limit is 2. The graph "jumps".f(x) = 1/x at x = 0. The limits go to +∞ and -∞.f(x) = sin(1/x) at x = 0. The function oscillates infinitely near 0, so no limit exists.A function f(x) is **differentiable** at a point x = a if its derivative exists at that point. The derivative, f'(a), represents the instantaneous rate of change, or the slope of the tangent line to the graph at that point.
Definition (First Principle):
f'(a) = lim (h→0) [f(a + h) - f(a)] / h
This limit must exist (and be a finite number).
This is a fundamental theorem in calculus.
Theorem: Differentiability implies Continuity.
If a functionf(x)is differentiable atx = a, then it *must* be continuous atx = a.
f(x) = |x| at x = 0.
This is the process of differentiating a function multiple times. The result of the first derivative is the *first derivative*, differentiating that gives the *second derivative*, and so on.
y' or f'(x) or dy/dxy'' or f''(x) or d²y/dx²y⁽ⁿ⁾ or f⁽ⁿ⁾(x) or dⁿy/dxⁿFinding a pattern for the n-th derivative is a common problem.
| Function f(x) | n-th Derivative f⁽ⁿ⁾(x) |
|---|---|
eᵃˣ |
aⁿ eᵃˣ |
aˣ |
(ln a)ⁿ aˣ |
xᵐ |
m(m-1)...(m-n+1) xᵐ⁻ⁿ = P(m, n) xᵐ⁻ⁿ |
sin(ax + b) |
aⁿ sin(ax + b + nπ/2) |
cos(ax + b) |
aⁿ cos(ax + b + nπ/2) |
log(x) |
(-1)ⁿ⁻¹ (n-1)! / xⁿ |
1 / (ax + b) |
(-1)ⁿ aⁿ n! / (ax + b)ⁿ⁺¹ |
This theorem provides a formula for the n-th derivative of a product of two functions. It is an extension of the product rule.
If y = u·v, where u and v are functions of x, then the n-th derivative of y is:
(uv)ₙ = C(n, 0)uₙv₀ + C(n, 1)uₙ₋₁v₁ + C(n, 2)uₙ₋₂v₂ + ... + C(n, r)uₙ₋ᵣvᵣ + ... + C(n, n)u₀vₙ
Where:
uₙ and vₙ represent the n-th derivatives of u and v.v₀ = v and u₀ = u (the original functions).C(n, r) = n! / (r! (n-r)!) is the binomial coefficient ("n choose r").Example: Find the 2nd derivative of y = x² sin(x) using Leibnitz's theorem.
Here n = 2. Let u = sin(x) and v = x² (Tip: Choose 'v' as the function that terminates to zero after a few derivatives).
u = sin(x), u₁ = cos(x), u₂ = -sin(x)
v = x², v₁ = 2x, v₂ = 2
y₂ = C(2, 0)u₂v₀ + C(2, 1)u₁v₁ + C(2, 2)u₀v₂
y₂ = (1)·(-sin(x))·(x²) + (2)·(cos(x))·(2x) + (1)·(sin(x))·(2)
y₂ = -x²sin(x) + 4x cos(x) + 2sin(x)
|x| example. Sharp corners break differentiability.x², x³) as your function 'v'. This is because its higher derivatives (v₂, v₃, ...) will become zero, making the formula much shorter.sin, cos, eᵃˣ, and 1/(ax+b). They are very common.