1. INTRODUCTION TO FUNCTION AND LIMIT
1.1 FUNCTION
A function is a relation where each input has exactly one output. The notation f(x) represents the output corresponding to input x.
Example:
f(x) = 5x squared plus 8
f(5) = 5 times 5 squared plus 8
f(5) = 125 plus 8
f(5) = 133
1.2 CONCEPT OF LIMIT
A limit is the value that a function or sequence approaches as the input or number of terms increases, without necessarily reaching it.
Example:
Sequence of line segment lengths: 12, 6, 3, 1.5, 0.75, 0.375, ...
Each time we divide the segment in half, the length decreases. As the number of divisions approaches infinity, the length approaches zero. Limit equals zero.
Thought Experiments:
Running half the distance to a finish line. You never reach exactly but get infinitely close.
Two people moving towards each other, each covering half the remaining distance. They never meet exactly but get very close.
2. LIMIT OF NUMBER SEQUENCE
2.1 GEOMETRIC SEQUENCE
A sequence in which each term is obtained by multiplying the previous term by a constant ratio.
General formula: a_n equals a_1 times r to the power n minus one
Example:
Sequence: 81, 27, 9, 3, ...
First term a_1 = 81
Common ratio r = 27 divided by 81 = 1 divided by 3
7th term: a_7 = 81 times (1/3)^6 = 1 divided by 9
Limit as n approaches infinity equals 0
2.2 EXERCISES
Sequence: 10, 1, 1/10, 1/100, ...
6th term = 0.00001
Limit = 0
Sequence: 5.01, 5.001, 5.0001, ...
Limit = 5
Divide a 1-foot line segment repeatedly in half
Lengths: 1, 1/2, 1/4, 1/8, ...
10th division: 1 divided by 512 approximately 0.001953 feet
3. LIMITS FROM GEOMETRIC FIGURES
3.1 REGULAR POLYGONS
Minimum sides for a closed polygon equals 3, which is an equilateral triangle
Sequence of regular polygons as sides increase: triangle, square, pentagon, hexagon, ...
Limit of a polygon with infinite sides equals a circle
3.2 FREQUENCY POLYGON
If the intervals of a histogram are made smaller, the number of segments increases and the polygon tends to a smooth curve.
Limiting value equals the frequency distribution curve
3.3 EXAMPLES
Triangle area division: triangle ABC has area 1 square unit. Midpoints form triangle PQR
Shaded area sequence: 1/4, 1/16, 1/64, ...
Sum of infinite series: S_infinity = (1/4) divided by (1 - 1/4) = 1/3
Concentric circles with radii r, r/2, r/4, ...
Areas: pi r squared, pi (r/2) squared, pi (r/4) squared, ...
Limit of area as n approaches infinity equals zero.
4. LIMIT AS SUM OF INFINITE SERIES
4.1 GEOMETRIC SERIES
Formula: S_infinity = a divided by (1 - r), where absolute value of r less than 1
Example: 1 + 1/2 + 1/4 + 1/8 + ...
S_infinity = 1 divided by (1 - 1/2) = 2
4.2 SHADED TRIANGLES
Sequence: 1/4 + 1/16 + 1/64 + ...
Sum: S_infinity = (1/4) divided by (1 - 1/4) = 1/3
5. LIMIT OF A FUNCTION
5.1 DEFINITION
If f(x) approaches L as x approaches a, then L is the limit of f(x) as x approaches a
Notation: limit as x approaches a of f(x) equals L
5.2 EXAMPLES
f(x) = 2x + 2, x approaches 2, f(x) approaches 6
f(x) = x squared minus 2, x approaches 3, f(x) approaches 7
f(x) = x cubed, x approaches 10, f(x) approaches 1000
f(x) = 1 divided by x squared, x approaches infinity, f(x) approaches 0
f(x) = x + 2, x approaches infinity, f(x) approaches infinity
6. SOLVED IMPORTANT QUESTIONS
Question 1
Find the 7th term of the sequence: 81, 27, 9, 3, ...
Solution:
Step 1: Identify the first term
a₁ = 81
Step 2: Find the common ratio
r = 27 ÷ 81 = 1/3
Step 3: Use the formula for the n-th term of a geometric sequence
aₙ = a₁ × r^(n-1)
Step 4: Substitute n = 7
a₇ = 81 × (1/3)^(7-1) = 81 × (1/3)^6
a₇ = 81 ÷ 729 = 1/9
Step 5: Limit as n → ∞
Since |r| < 1, the sequence approaches 0.
Answer: a₇ = 1/9, Limit = 0
Question 2
Find the 6th term of the sequence: 10, 1, 1/10, 1/100, ...
Solution:
Step 1: First term a₁ = 10
Step 2: Common ratio r = 1 ÷ 10 = 1/10
Step 3: Use n-th term formula: a₆ = a₁ × r^(6-1)
a₆ = 10 × (1/10)^5 = 10 × 1/100000 = 0.0001
Step 4: Limit as n → ∞
Since |r| < 1, the sequence approaches 0
Answer: a₆ = 0.0001, Limit = 0
Question 3
Sequence: 5.01, 5.001, 5.0001, ...
Find the limit as n → ∞
Solution:
Observe that each term is getting closer to 5.
So, as n → ∞, the sequence approaches 5
Answer: Limit = 5
Question 4
Divide a 1-foot segment repeatedly in half. Find the length at 10th division.
Solution:
Length after each division forms a geometric sequence:
1, 1/2, 1/4, 1/8, ...
Formula for n-th term: aₙ = a₁ × (1/2)^(n-1)
a₁ = 1, n = 10
a₁₀ = 1 × (1/2)^(10-1) = 1 × (1/2)^9 = 1/512 ≈ 0.001953 feet
Answer: 1/512 feet ≈ 0.001953 feet
Question 5
Minimum sides of a polygon and limit of infinite-sided polygon
Solution:
Minimum number of sides for a polygon = 3 → Triangle
As the number of sides increases infinitely → Polygon approaches a circle
Answer: Minimum sides = 3, Limit = Circle
Question 6
Triangle ABC area = 1 sq unit. Midpoints form triangle PQR. Find sum of shaded areas.
Solution:
Step 1: First shaded triangle area = 1/4 of ABC = 1 × 1/4 = 1/4
Step 2: Second shaded triangle area = 1/4 of previous = 1/16
Step 3: Third shaded triangle area = 1/4 of previous = 1/64
Step 4: Infinite series: S = 1/4 + 1/16 + 1/64 + ...
S∞ = a / (1 - r) = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3
Answer: Sum of shaded areas = 1/3
Question 7
Concentric circles with radii r, r/2, r/4, ... Find the limiting area
Solution:
Area sequence: π r², π (r/2)² = π r²/4, π (r/4)² = π r²/16, ...
Infinite series sum: S∞ = π r² / (1 - 1/4) = π r² / (3/4) → Actually, we want the limiting smallest circle → as radius → 0, area → 0
Answer: Limit of area = 0
Question 8
Sum of series: 1 + 1/2 + 1/4 + 1/8 + ...
Solution:
Geometric series with first term a = 1, r = 1/2
Sum to infinity: S∞ = a / (1 - r) = 1 / (1 - 1/2) = 1 / (1/2) = 2
Answer: Sum = 2
Question 9
Shaded triangle series: 1/4 + 1/16 + 1/64 + ...
Solution:
Geometric series with a = 1/4, r = 1/4
Sum to infinity: S∞ = a / (1 - r) = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3
Answer: Sum = 1/3
Question 10
f(x) = 2x + 2, limit as x → 2
Solution:
f(2) = 2 × 2 + 2 = 6
Answer: Limit = 6
Question 11
f(x) = x² − 2, limit as x → 3
Solution:
f(3) = 3² − 2 = 9 − 2 = 7
Answer: Limit = 7
Question 12
f(x) = x³, limit as x → 10
Solution:
f(10) = 10³ = 1000
Answer: Limit = 1000
Question 13
f(x) = 1 / x², limit as x → ∞
Solution:
As x increases indefinitely, 1 / x² → 0
Answer: Limit = 0
Question 14
f(x) = x + 2, limit as x → ∞
Solution:
As x increases indefinitely, x + 2 → ∞
Answer: Limit = ∞
Question 15
Running half the distance to exit
Solution:
Every step covers half the remaining distance.
You will never reach exactly the exit, but get infinitely close.
This demonstrates a real-life example of limit.
Answer: Never reaches exactly, approaches the exit
Question 16
Divide Rs. 2,621,440 repeatedly by 2. Find remaining after 20th division
Solution:
Step 1: First division: 2,621,440 / 2
Step 2: After n divisions: Remaining = 2,621,440 / 2^20
Step 3: 2^20 = 1,048,576
Remaining ≈ 2.5
Answer: Remaining ≈ 2.5
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