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Calculus - Limits and Convergence

Grade 11IB

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

🔑Concepts

Intuition of a Limit: A limit describes the value that a function f(x)f(x) approaches as the input xx gets closer and closer to a specific point aa. Visually, this is represented by following the curve of a graph from both the left and right sides toward x=ax = a to see if the yy-values converge on a single height LL, even if there is a 'hole' or open circle at that exact point.

One-Sided Limits: A limit can be approached from the left (xax \to a^-) or the right (xa+x \to a^+). Visually, if you trace the graph from the left and it ends at height L1L_1, but tracing from the right ends at height L2L_2, and L1L2L_1 \neq L_2, the graph exhibits a 'jump discontinuity' and the general limit does not exist.

Existence of a Limit: For a limit limxaf(x)\lim_{x \to a} f(x) to exist, the left-hand limit and the right-hand limit must be equal. On a graph, this means the two 'branches' of the function must meet at the same vertical level at x=ax = a.

Continuity: A function is continuous at a point x=cx = c if three conditions are met: f(c)f(c) is defined, the limit as xcx \to c exists, and the limit equals the function value. Visually, a continuous function can be drawn without lifting your pencil, showing no breaks, jumps, or vertical asymptotes.

Limits at Infinity and Asymptotes: These describe the 'end behavior' of a function. If limxf(x)=L\lim_{x \to \infty} f(x) = L, the graph approaches a horizontal line y=Ly = L, known as a horizontal asymptote. If the function value grows without bound (x±x \to \pm \infty) as xx approaches a finite value cc, the graph features a vertical dashed line called a vertical asymptote.

Indeterminate Forms: When substituting a value results in 00\frac{0}{0} or \frac{\infty}{\infty}, the limit cannot be determined immediately. This usually indicates a 'hole' in the graph. Algebraically, this is resolved by factoring, rationalizing, or simplifying to reveal the limit's true value.

Convergence of Sequences and Series: A sequence converges if its terms approach a finite limit LL as nn \to \infty. For a geometric series, convergence only occurs if the common ratio r<1|r| < 1. Visually, the sum of a convergent series approaches a horizontal 'ceiling' or limit, while a divergent series grows infinitely or oscillates.

📐Formulae

Definition of a limit: limxaf(x)=L\lim_{x \to a} f(x) = L

Existence condition: limxaf(x)=limxa+f(x)=L\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L

Condition for continuity: limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) sentiments

Sum to infinity of a geometric series: S=u11rS_{\infty} = \frac{u_1}{1 - r} for r<1|r| < 1

Limit of a constant: limxak=k\lim_{x \to a} k = k

Power rule for limits: limxa[f(x)]n=[limxaf(x)]n\lim_{x \to a} [f(x)]^n = [\lim_{x \to a} f(x)]^n

Rational function limit at infinity: limx1xn=0\lim_{x \to \infty} \frac{1}{x^n} = 0 for n>0n > 0

💡Examples

Problem 1:

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

Solution:

Step 1: Attempt direct substitution. 32933=00\frac{3^2 - 9}{3 - 3} = \frac{0}{0}. This is an indeterminate form. Step 2: Factor the numerator. x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3). Step 3: Simplify the expression. (x3)(x+3)x3=x+3\frac{(x - 3)(x + 3)}{x - 3} = x + 3 for x3x \neq 3. Step 4: Re-evaluate the limit. limx3(x+3)=3+3=6\lim_{x \to 3} (x + 3) = 3 + 3 = 6.

Explanation:

Since direct substitution led to 00\frac{0}{0}, we used algebraic factoring to remove the common factor causing the zero in the denominator, revealing the limit value at the 'hole' in the graph.

Problem 2:

Determine if the infinite geometric series 10+5+2.5+...10 + 5 + 2.5 + ... converges, and if so, find its sum.

Solution:

Step 1: Identify the first term u1u_1 and the common ratio rr. Here, u1=10u_1 = 10 and r=510=0.5r = \frac{5}{10} = 0.5. Step 2: Check for convergence. Since r=0.5<1|r| = |0.5| < 1, the series converges. Step 3: Use the sum to infinity formula. S=u11rS_{\infty} = \frac{u_1}{1 - r}. Step 4: Substitute the values. S=1010.5=100.5=20S_{\infty} = \frac{10}{1 - 0.5} = \frac{10}{0.5} = 20.

Explanation:

Because the terms get smaller at a constant rate where the ratio is less than 1, the total sum approaches a finite limit rather than growing to infinity.