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Limits in Calculus

Limits describe the value a function approaches as the input approaches some point.

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

The foundation of calculus, enabling concepts like derivatives and integrals.

Why Limits Matter

Limits help us understand instantaneous change and the behavior of functions near specific points.

They allow us to analyze functions at points where they might not be defined.

Notation

Limit of f(x) as x approaches a is written as:

\( \lim_{x \to a} f(x) \)

We read this as "the limit of f(x) as x approaches a"

Examples of Limits

Simple substitution example:

\( \lim_{x \to 2} (3x+1) = 7 \)
Step 1: Substitute x = 2 into the function
Step 2: 3(2) + 1 = 6 + 1 = 7

Types of Limits

One-sided limits (left and right), two-sided limits, infinite limits, and limits at infinity.

\( \lim_{x \to a^-} f(x) \quad \text{vs} \quad \lim_{x \to a^+} f(x) \)

Substitution Method

Directly substitute the value into the function when it does not create indeterminate forms.

This works when the function is continuous at the point.

Example 1: Substitution

\( \lim_{x \to 3} (2x+5) \)
Step 1: Substitute x = 3
Step 2: 2(3) + 5 = 6 + 5 = 11
Answer: \( \lim_{x \to 3} (2x+5) = 11 \)

Example 2: Substitution

\( \lim_{x \to -1} (x^2+4) \)
Step 1: Substitute x = -1
Step 2: (-1)² + 4 = 1 + 4 = 5
Answer: \( \lim_{x \to -1} (x^2+4) = 5 \)

Example 3: Substitution

\( \lim_{x \to 0} (5x-3) \)
Step 1: Substitute x = 0
Step 2: 5(0) - 3 = 0 - 3 = -3
Answer: \( \lim_{x \to 0} (5x-3) = -3 \)

Key Tip for Substitution

Only use substitution when the function is continuous at the point.

If you get an indeterminate form (like 0/0), you need a different approach.

Practice 1: Substitution

\( \lim_{x \to 4} (x^2-1) \)
Step 1: Substitute x = 4
Step 2: (4)² - 1 = 16 - 1 = 15
Answer: \( \lim_{x \to 4} (x^2-1) = 15 \)

Practice 2: Substitution

\( \lim_{x \to -2} (3x^2+1) \)
Step 1: Substitute x = -2
Step 2: 3(-2)² + 1 = 3(4) + 1 = 12 + 1 = 13
Answer: \( \lim_{x \to -2} (3x^2+1) = 13 \)

Practice 3: Substitution

\( \lim_{x \to 1} (x^3-2x+4) \)
Step 1: Substitute x = 1
Step 2: (1)³ - 2(1) + 4 = 1 - 2 + 4 = 3
Answer: \( \lim_{x \to 1} (x^3-2x+4) = 3 \)

Substitution Summary

Directly substitute the value when function is continuous and defined at the point.

If substitution gives a valid number, that's the limit!

Next Method: Factorization

Used when substitution gives 0/0 indeterminate form.

Factor and cancel common terms to resolve the indeterminacy.

Factorization Method

Factor the numerator and/or denominator to simplify limits.

Particularly useful for rational functions with common factors.

Example 1: Factorization

\( \lim_{x \to 2} \frac{x^2-4}{x-2} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Factor the numerator: x² - 4 = (x-2)(x+2)
Step 3: Cancel common factor: \( \frac{(x-2)(x+2)}{x-2} = x+2 \)
Step 4: Now substitute x = 2: 2 + 2 = 4
Answer: \( \lim_{x \to 2} \frac{x^2-4}{x-2} = 4 \)

Example 2: Factorization

\( \lim_{x \to 3} \frac{x^2-9}{x-3} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Factor the numerator: x² - 9 = (x-3)(x+3)
Step 3: Cancel common factor: \( \frac{(x-3)(x+3)}{x-3} = x+3 \)
Step 4: Now substitute x = 3: 3 + 3 = 6
Answer: \( \lim_{x \to 3} \frac{x^2-9}{x-3} = 6 \)

Key Tip for Factorization

Always factor completely and cancel common terms before substituting.

Look for difference of squares, perfect squares, and other factoring patterns.

Practice 1: Factorization

\( \lim_{x \to 5} \frac{x^2-25}{x-5} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Factor the numerator: x² - 25 = (x-5)(x+5)
Step 3: Cancel common factor: \( \frac{(x-5)(x+5)}{x-5} = x+5 \)
Step 4: Now substitute x = 5: 5 + 5 = 10
Answer: \( \lim_{x \to 5} \frac{x^2-25}{x-5} = 10 \)

Practice 2: Factorization

\( \lim_{x \to 4} \frac{x^2-16}{x-4} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Factor the numerator: x² - 16 = (x-4)(x+4)
Step 3: Cancel common factor: \( \frac{(x-4)(x+4)}{x-4} = x+4 \)
Step 4: Now substitute x = 4: 4 + 4 = 8
Answer: \( \lim_{x \to 4} \frac{x^2-16}{x-4} = 8 \)

Practice 3: Factorization

\( \lim_{x \to 1} \frac{x^2-1}{x-1} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Factor the numerator: x² - 1 = (x-1)(x+1)
Step 3: Cancel common factor: \( \frac{(x-1)(x+1)}{x-1} = x+1 \)
Step 4: Now substitute x = 1: 1 + 1 = 2
Answer: \( \lim_{x \to 1} \frac{x^2-1}{x-1} = 2 \)

Factorization Summary

Factorization is used to remove 0/0 indeterminate forms and simplify the limit.

Remember to factor completely before canceling terms.

Next Method: Trigonometry

For limits involving sine, cosine, or other trig functions.

Uses special trigonometric limits and identities.

Trigonometric Limits

Use trig identities or standard limits to evaluate these.

Two especially important limits form the foundation:

Standard Limit 1

\( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

This is a fundamental limit used throughout calculus.

Standard Limit 2

\( \lim_{x \to 0} \frac{1-\cos x}{x} = 0 \)

Another important limit for trigonometric functions.

Example: Trigonometric Limit

\( \lim_{x \to 0} \frac{\sin 3x}{x} \)
Step 1: Multiply numerator and denominator by 3: \( \frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \)
Step 2: As x → 0, 3x → 0, so \( \frac{\sin 3x}{3x} \to 1 \)
Step 3: Therefore, the limit is 3 × 1 = 3
Answer: \( \lim_{x \to 0} \frac{\sin 3x}{x} = 3 \)

Tip for Trigonometric Limits

Use identities like \( \sin^2 x + \cos^2 x = 1 \) when necessary.

For limits involving tan x, remember that \( \tan x = \frac{\sin x}{\cos x} \).

Practice 1: Trigonometric Limit

\( \lim_{x \to 0} \frac{\tan x}{x} \)
Step 1: Write tan x as sin x / cos x: \( \frac{\tan x}{x} = \frac{\sin x}{x \cos x} \)
Step 2: Separate: \( \frac{\sin x}{x} \cdot \frac{1}{\cos x} \)
Step 3: As x → 0, \( \frac{\sin x}{x} \to 1 \) and \( \frac{1}{\cos x} \to \frac{1}{1} = 1 \)
Step 4: Therefore, the limit is 1 × 1 = 1
Answer: \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

Practice 2: Trigonometric Limit

\( \lim_{x \to 0} \frac{\sin 2x}{x} \)
Step 1: Multiply numerator and denominator by 2: \( \frac{\sin 2x}{x} = 2 \cdot \frac{\sin 2x}{2x} \)
Step 2: As x → 0, 2x → 0, so \( \frac{\sin 2x}{2x} \to 1 \)
Step 3: Therefore, the limit is 2 × 1 = 2
Answer: \( \lim_{x \to 0} \frac{\sin 2x}{x} = 2 \)

Practice 3: Trigonometric Limit

\( \lim_{x \to 0} \frac{1-\cos 2x}{x} \)
Step 1: Use the identity: 1 - cos 2x = 2 sin² x
Step 2: Rewrite the limit: \( \lim_{x \to 0} \frac{2 \sin^2 x}{x} = 2 \lim_{x \to 0} \frac{\sin x}{x} \cdot \sin x \)
Step 3: As x → 0, \( \frac{\sin x}{x} \to 1 \) and sin x → 0
Step 4: Therefore, the limit is 2 × 1 × 0 = 0
Answer: \( \lim_{x \to 0} \frac{1-\cos 2x}{x} = 0 \)

Trigonometric Limits Summary

Trigonometric limits often require identities or standard limits.

Remember the two fundamental limits: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{1-\cos x}{x} = 0 \)

Next Method: Rationalization

Used when surds (square roots) appear in limits.

Multiply numerator and denominator by the conjugate to eliminate radicals.

Rationalization Method

Multiply numerator and denominator by the conjugate to remove square roots.

Particularly useful for limits that result in 0/0 when direct substitution is used.

Example 1: Rationalization

\( \lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply numerator and denominator by the conjugate: \( \frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} \)
Step 3: Numerator becomes: (√x - 2)(√x + 2) = x - 4
Step 4: Denominator becomes: (x - 4)(√x + 2)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+2} \)
Step 6: Now substitute x = 4: \( \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4} \)
Answer: \( \lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} = \frac{1}{4} \)

Example 2: Rationalization

\( \lim_{x \to 9} \frac{\sqrt{x}-3}{x-9} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply numerator and denominator by the conjugate: \( \frac{\sqrt{x}-3}{x-9} \cdot \frac{\sqrt{x}+3}{\sqrt{x}+3} \)
Step 3: Numerator becomes: (√x - 3)(√x + 3) = x - 9
Step 4: Denominator becomes: (x - 9)(√x + 3)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+3} \)
Step 6: Now substitute x = 9: \( \frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6} \)
Answer: \( \lim_{x \to 9} \frac{\sqrt{x}-3}{x-9} = \frac{1}{6} \)

Example 3: Rationalization

\( \lim_{x \to 1} \frac{\sqrt{x+3}-2}{x-1} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply numerator and denominator by the conjugate: \( \frac{\sqrt{x+3}-2}{x-1} \cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} \)
Step 3: Numerator becomes: (√(x+3) - 2)(√(x+3) + 2) = (x+3) - 4 = x - 1
Step 4: Denominator becomes: (x - 1)(√(x+3) + 2)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x+3}+2} \)
Step 6: Now substitute x = 1: \( \frac{1}{\sqrt{1+3}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4} \)
Answer: \( \lim_{x \to 1} \frac{\sqrt{x+3}-2}{x-1} = \frac{1}{4} \)

Practice 1: Rationalization

\( \lim_{x \to 16} \frac{\sqrt{x}-4}{x-16} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply by conjugate: \( \frac{\sqrt{x}-4}{x-16} \cdot \frac{\sqrt{x}+4}{\sqrt{x}+4} \)
Step 3: Numerator becomes: (√x - 4)(√x + 4) = x - 16
Step 4: Denominator becomes: (x - 16)(√x + 4)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+4} \)
Step 6: Now substitute x = 16: \( \frac{1}{\sqrt{16}+4} = \frac{1}{4+4} = \frac{1}{8} \)
Answer: \( \lim_{x \to 16} \frac{\sqrt{x}-4}{x-16} = \frac{1}{8} \)

Practice 2: Rationalization

\( \lim_{x \to 25} \frac{\sqrt{x}-5}{x-25} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply by conjugate: \( \frac{\sqrt{x}-5}{x-25} \cdot \frac{\sqrt{x}+5}{\sqrt{x}+5} \)
Step 3: Numerator becomes: (√x - 5)(√x + 5) = x - 25
Step 4: Denominator becomes: (x - 25)(√x + 5)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+5} \)
Step 6: Now substitute x = 25: \( \frac{1}{\sqrt{25}+5} = \frac{1}{5+5} = \frac{1}{10} \)
Answer: \( \lim_{x \to 25} \frac{\sqrt{x}-5}{x-25} = \frac{1}{10} \)

Practice 3: Rationalization

\( \lim_{x \to 36} \frac{\sqrt{x}-6}{x-36} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply by conjugate: \( \frac{\sqrt{x}-6}{x-36} \cdot \frac{\sqrt{x}+6}{\sqrt{x}+6} \)
Step 3: Numerator becomes: (√x - 6)(√x + 6) = x - 36
Step 4: Denominator becomes: (x - 36)(√x + 6)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+6} \)
Step 6: Now substitute x = 36: \( \frac{1}{\sqrt{36}+6} = \frac{1}{6+6} = \frac{1}{12} \)
Answer: \( \lim_{x \to 36} \frac{\sqrt{x}-6}{x-36} = \frac{1}{12} \)

Tip for Rationalization

Always multiply by conjugate to simplify expressions with radicals.

The conjugate of √a - b is √a + b, and vice versa.

Practice 4: Rationalization

\( \lim_{x \to 49} \frac{\sqrt{x}-7}{x-49} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply by conjugate: \( \frac{\sqrt{x}-7}{x-49} \cdot \frac{\sqrt{x}+7}{\sqrt{x}+7} \)
Step 3: Numerator becomes: (√x - 7)(√x + 7) = x - 49
Step 4: Denominator becomes: (x - 49)(√x + 7)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+7} \)
Step 6: Now substitute x = 49: \( \frac{1}{\sqrt{49}+7} = \frac{1}{7+7} = \frac{1}{14} \)
Answer: \( \lim_{x \to 49} \frac{\sqrt{x}-7}{x-49} = \frac{1}{14} \)

Practice 5: Rationalization

\( \lim_{x \to 64} \frac{\sqrt{x}-8}{x-64} \)
Step 1: Direct substitution gives 0/0 (indeterminate)
Step 2: Multiply by conjugate: \( \frac{\sqrt{x}-8}{x-64} \cdot \frac{\sqrt{x}+8}{\sqrt{x}+8} \)
Step 3: Numerator becomes: (√x - 8)(√x + 8) = x - 64
Step 4: Denominator becomes: (x - 64)(√x + 8)
Step 5: Cancel common factor: \( \frac{1}{\sqrt{x}+8} \)
Step 6: Now substitute x = 64: \( \frac{1}{\sqrt{64}+8} = \frac{1}{8+8} = \frac{1}{16} \)
Answer: \( \lim_{x \to 64} \frac{\sqrt{x}-8}{x-64} = \frac{1}{16} \)

Rationalization Summary

Rationalization removes surds and helps evaluate limits in indeterminate forms.

Multiply numerator and denominator by the conjugate, then simplify.

Congratulations!

You have completed your first lesson on Limits in Calculus.

You've learned multiple techniques with detailed examples: