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Calculus - Limits, Continuity and Differentiation from First Principles

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Limit Definition: A limit limxaf(x)=L\lim_{x \to a} f(x) = L describes the value f(x)f(x) approaches as xx gets arbitrarily close to aa from both sides. On a graph, imagine a point moving along the curve from the left and another from the right; if they both target the same yy-value height LL, the limit exists. Note that the function does not need to be defined at x=ax=a for the limit to exist.

One-Sided Limits: Limits can be evaluated from the left (denoted xax \to a^-) or the right (denoted xa+x \to a^+). Visually, these represent the paths taken toward a specific x-coordinate from different directions. For a two-sided limit to exist, these two paths must converge at the exact same vertical point.

Continuity: A function is continuous at x=ax=a if three conditions are satisfied: f(a)f(a) is defined, limxaf(x)\lim_{x \to a} f(x) exists, and limxaf(x)=f(a)\lim_{x \to a} f(x) = f(a). On a graph, this appears as an unbroken line or curve. If you have to lift your pencil to draw the function, it is discontinuous at that point (due to a hole, jump, or asymptote).

Removable vs. Non-Removable Discontinuities: A removable discontinuity occurs when a limit exists but is not equal to the function's value, visually appearing as a single 'hole' in the line. A non-removable discontinuity, like a vertical asymptote or a jump, occurs where the left-hand and right-hand limits do not match or where the function goes to infinity.

Gradient of a Secant Line: The average rate of change between two points (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)) is represented by the slope of the secant line. Geometrically, this is a straight line cutting through the curve at two distinct points. The formula for this gradient is f(x+h)f(x)h\frac{f(x+h) - f(x)}{h}.

Differentiation from First Principles: The derivative f(x)f'(x) is the instantaneous rate of change or the gradient of the tangent line at a specific point. It is found by taking the limit of the secant gradient as the horizontal distance hh between the two points shrinks to zero. Graphically, as h0h \to 0, the secant line rotates until it just touches the curve at a single point, becoming the tangent.

Tangent Line Properties: The tangent line at a point (a,f(a))(a, f(a)) has a gradient equal to f(a)f'(a). It represents the 'slope' of the curve at that exact moment. If f(a)>0f'(a) > 0, the graph is rising at that point; if f(a)<0f'(a) < 0, it is falling; and if f(a)=0f'(a) = 0, the graph is momentarily flat (a horizontal tangent).

📐Formulae

limxaf(x)=L\lim_{x \to a} f(x) = L

f(a)=limxaf(x)=limxa+f(x)f(a) = \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) (Condition for Continuity)

msecant=f(x+h)f(x)hm_{secant} = \frac{f(x+h) - f(x)}{h}

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} (First Principles)

yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a) (Equation of Tangent Line)

💡Examples

Problem 1:

Evaluate the limit: limx3x29x3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Solution:

  1. Attempt direct substitution: 32933=00\frac{3^2 - 9}{3 - 3} = \frac{0}{0}. This is an indeterminate form.
  2. Factor the numerator using the difference of squares: x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3).
  3. Rewrite the limit: limx3(x3)(x+3)x3\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}.
  4. Cancel the common factor (x3)(x - 3): limx3(x+3)\lim_{x \to 3} (x + 3).
  5. Substitute x=3x = 3: 3+3=63 + 3 = 6.

Explanation:

The function has a removable discontinuity (a hole) at x=3x=3. By factoring and simplifying, we can find the value the function approaches even though it is undefined at that specific point.

Problem 2:

Find the derivative of f(x)=x2+xf(x) = x^2 + x from first principles.

Solution:

  1. State the formula: f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
  2. Substitute the function: f(x)=limh0[(x+h)2+(x+h)][x2+x]hf'(x) = \lim_{h \to 0} \frac{[(x+h)^2 + (x+h)] - [x^2 + x]}{h}.
  3. Expand the terms: f(x)=limh0x2+2xh+h2+x+hx2xhf'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + x + h - x^2 - x}{h}.
  4. Simplify the numerator: f(x)=limh02xh+h2+hhf'(x) = \lim_{h \to 0} \frac{2xh + h^2 + h}{h}.
  5. Factor out hh from the numerator: f(x)=limh0h(2x+h+1)hf'(x) = \lim_{h \to 0} \frac{h(2x + h + 1)}{h}.
  6. Divide by hh: f(x)=limh0(2x+h+1)f'(x) = \lim_{h \to 0} (2x + h + 1).
  7. Apply the limit as h0h \to 0: f(x)=2x+0+1=2x+1f'(x) = 2x + 0 + 1 = 2x + 1.

Explanation:

We use the algebraic definition of a derivative. By expanding and simplifying, we remove the hh in the denominator, allowing us to evaluate the limit without dividing by zero.